Answer:
See explaination
Explanation:
Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,
AgF + NaCl= AgCl + NaF
Here, the color of AgCl is white, hence the solution cannot be AgCl.
Determination of NaCl
Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.
Reactions
Ag+ (aq)+ Cl-(aq) = AgCl(aq)
Ag+(aq) + SCN-(aq) = AgSCN(aq)
Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)
EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pKa3=2.00 , p K a4 = 2.69 pKa4=2.69 , p K a5 = 6.13 pKa5=6.13 , and p K a6 = 10.37 pKa6=10.37 . The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA EDTA , it is convenient to calculate the fraction of EDTA EDTA that is in the completely unprotonated form, Y 4 − Y4− . This fraction is designated α Y 4 − αY4− . Calculate α Y 4 − αY4− at two pH values.
To calculate αY4-, we need to find the fraction of EDTA in its protonated forms at the given pH and divide the concentration of the completely unprotonated form by the total concentration of EDTA.
Explanation:To calculate the fraction of EDTA that is in the completely unprotonated form, Y4- (designated as αY4-), at a given pH value, we need to find the values of αY4- at two different pH values.
To calculate αY4-, we first need to find the fraction of EDTA in its protonated forms at the given pH.
For example, at pH = 2, we can calculate the fractional composition of each protonated form by dividing the concentration of the protonated form by the total concentration of all protonation states of EDTA. We can repeat this calculation for another pH value to find αY4- at that pH.
Learn more about EDTA protonation equilibrium here:https://brainly.com/question/16232441
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The fraction of EDTA in the completely unprotonated form at pH3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].
To calculate the fraction of EDTA that is in the completely unprotonated form, [tex]\( \alpha_{Y^{4-}} \)[/tex], at a given pH, we use the following equation:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + \sum_{i=1}^{6} 10^{(pH - pKa_i)}} \][/tex]
Here, [tex]\( pKa_i \)[/tex] represents the [tex]\( i \)-th[/tex] acid dissociation constant for EDTA. Since EDTA is a hexaprotic acid, it has six [tex]\( pKa \)[/tex] values, which are given as:
[tex]\( pKa_1 = 0.00 \), \( pKa_2 = 1.50 \), \( pKa_3 = 2.00 \), \( pKa_4 = 2.69 \), \( pKa_5 = 6.13 \), and \( pKa_6 = 10.37 \).[/tex]
Calculate [tex]\( \alpha_{Y^{4-}} \)[/tex] for two pH values, say pH = 3 and pH = 8.
For pH = 3:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(3 - 0.00)} + 10^{(3 - 1.50)} + 10^{(3 - 2.00)} + 10^{(3 - 2.69)} + 10^{(3 - 6.13)} + 10^{(3 - 10.37)}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{3} + 10^{1.5} + 10^{1} + 10^{0.31} + 10^{-3.13} + 10^{-7.37}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 1000 + 31.62 + 10 + 2.08 + 0.00074 + 4.17 \times 10^{-8}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1 + 1000 + 31.62 + 10 + 2.08} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1033.62 + 31.62 + 10 + 2.08} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1077.32} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx 9.28 \times 10^{-4} \][/tex]
For pH = 8:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(8 - 0.00)} + 10^{(8 - 1.50)} + 10^{(8 - 2.00)} + 10^{(8 - 2.69)} + 10^{(8 - 6.13)} + 10^{(8 - 10.37)}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{8} + 10^{6.5} + 10^{6} + 10^{5.31} + 10^{1.87} + 10^{-2.37}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10000000 + 316227.76 + 1000000 + 20794.42 + 74.12 + 0.00417} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{10000000 + 316227.76 + 1000000 + 20794.42 + 74.12} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1131992.3} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx 8.83 \times 10^{-7} \][/tex]
Therefore, the fraction of EDTA in the completely unprotonated form at pH 3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH 8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].
The complete question is:
EDTA is a hexaprotic system with the PK, values:
PKa1 = 0.00, pKa2 = 1.50, pKa3 2.00, pKa4 = 2.69,
pKa5 6.13, and pKa6 10.37.
The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, Y4-. This fraction is designated αY4−. Calculate αY4− at two pH values.
pH = 3.10
pH = 10.55
Which of the following statements concerning hydrocarbons is/are correct?
1.
Saturated hydrocarbons may be cyclic or acyclic molecules.
2.
An unsaturated hydrocarbon molecule contains at least one double bond.
3.
Ethylenediamine, H2NCH2CH2NH2, is an example of a saturated hydrocarbon.
Answer:
1. Saturated hydrocarbons may be cyclic or acyclic molecules.
2. An unsaturated hydrocarbon molecule contains at least one double bond.
Explanation:
Hello,
In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).
Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true
Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.
Therefore, first and second statements are correct.
Best regards.
Final answer:
Statements 1 and 2 are correct regarding saturated hydrocarbons being cyclic or acyclic and unsaturated hydrocarbons containing at least one double bond. Statement 3 is incorrect because ethylenediamine is not a hydrocarbon.
Explanation:
To address which statements concerning hydrocarbons are correct:
Saturated hydrocarbons may indeed be cyclic or acyclic molecules. When cyclic, they have the general formula CnH₂n, such as cycloalkanes, which are saturated compounds. Acyclic saturated hydrocarbons, also known as alkanes, have single bonds only and follow the general formula CHn₂+2.
An unsaturated hydrocarbon molecule contains at least one double or triple bond. Molecules with one or more double bonds are alkenes, with the simplest being ethene (or ethylene), C₂H₄.
Ethylenediamine, H₂NCH₂CH₂NH₂, is not an example of a saturated hydrocarbon. It contains amine groups (NH₂) and therefore is not a hydrocarbon.
Hence, statements 1 and 2 are correct, while statement 3 is incorrect.
Using the appropriate Ksp values, find the concentration of K+ ions in the solution at equilibrium after 600 mL of 0.45 M aqueous Cu(NO3)2 solution has been mixed with 450 mL of 0.25 M aqueous KOH solution. (Enter in M.) (Ksp for Cu(OH)2 is 2.6x10-19).
Now find the concentration of OH? ions in this solution at equilibrium.
Answer:
[K⁺] = 0.107 M
[OH⁻] = 1.13 × 10⁻⁹ M
Explanation:
600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻
⇒ 0.45 × 600 × 10⁻³
= 0.27 moles of Cu²⁺ and (NO₃)²⁻
450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻
⇒ 0.25 × 450 × 10⁻³
= 0.1125 moles of K⁺ and OH⁻
Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺ (because 1 Cu²⁺ needs 2 OH⁻)
Therefore , moles of remaining Cu²⁺ = 0.27 - 0.05625
=0.21375 moles which is equal to :
⇒ 0.21375/(( 600+450))× 10⁻³
= 0.21375/1050 × 10⁻³
= 0.20357 M
Given that :
(Ksp for Cu(OH)2 is 2.6 × 10⁻¹⁹)
We know that , Ksp = [Cu²⁺][OH⁻]²
2.6 × 10⁻¹⁹ = 0.20357 × [OH⁻]²
[OH⁻]² = 2.6 × 10⁻¹⁹/0.20357
[OH⁻] = 1.13 × 10⁻⁹ M
[K⁺] = moles of K⁺ /total volume
[K⁺] = 0.1125 / 1050 × 10⁻³
[K⁺] = 0.107 M
Consider the following reaction.
CO2(g) + H2(9)
-
CO(g) + H2O(1)
What is being oxidized?
O carbon
carbon dioxide
oxygen
o hydrogen
Answer:
Hydrogen
Explanation:
It gains oxygen from carbon dioxide to form water
Oxidation is the addition of oxygen to an element in a chemical reaction
Answer:
Its D
Explanation:
Trust me just get ronas and cheat
Retigeranic Acid II: The Endgame. This question deals with transformations that were employed in the last few steps of a total synthesis of Retigeranic Acid. (a) Provide reagents that will accomplish the transformation of molecule 6 to molecule 7. (b) Draw the mechanism of the transformation of molecule 7 to molecule 8 when treated with sodium hydroxide in water. (It is ok to abbreviate parts of the molecule not involved in the reaction to make drawing a little easier).
Answer:
see explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
which molecule is a product of photosynthesis?
Carbon dioxide
Sugar
Water
Nitrogen
Consider an electrochemical cell based on the spontaneous reaction 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+. If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should A) increase by 0.06 V. D) decrease by 0.18 V. B) increase by 0.18 V. E) increase by 0.35 V. C) decrease by 0.06 V.
Answer:
B) increase by 0.18 V
Explanation:
The given chemical spontaneous reaction is :
[tex]2 AgCl_{(s)} + Zn _{(s)} \to 2Ag (s) + 2Cl^- _{(aq)} + Zn ^{2+} _{(aq)}[/tex]
By applying Nernst Equation:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
here ;
n = number of electrons transferred in the reaction
n =2
[tex]E^0 = E^0_{cathode } - E^0_{anode}[/tex]
[tex]E^0 = E^0_{Ag^+/Ag } - E^0_{Zn^{2+}/Zn}[/tex]
[tex]E^0 =+0.80 \ V -(-0.76 \ V)[/tex]
[tex]E^0 =1.56 \ V[/tex]
When it happens to occur that the concentration of chlorine (aq) and Zn²⁺ (aq) is 1 M ;
[tex]E^0_{cell} = E^0[/tex] is as follows:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1) \ \ \ \ \ \ \ ( where \ log (1) = 0)[/tex]
[tex]E_{cell} = 1.56 \ V[/tex]
Now; the [tex]E_{cell}[/tex] value in the decreased concentration of chlorine (aq) ion is calculated as:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1*0.001^2)[/tex]
[tex]E_{cell} = +1.737 \ V[/tex]
Hence; the change in voltage = [tex]E_{cell} - E^0[/tex]
= 1.737 - 1.56
= 0.177 V
≅ 0.18 V
We therefore conclude that: since the [tex]E^0_{cell}[/tex] value after the decreased concentration of Chlorine is greater than the [tex]E^0[/tex] before the change; then there is increase in the value by 0.18 V
_ H₂ + O2
H20
balance equation
Answer:
2H2 + O2 = 2H2O
Explanation:
From the original equation, you first need to write each component separately.
Left side: H = 2 ; O = 2 (number based on the subscript)
Right side: H = 2 ; O = 1 (number based on the subscripts; no subscript means that the element is just 1)
Notice that the number of H is already balance but the number of O is not. In order to balance the O, you need to multiply the element by 2, but you CANNOT do this by simply changing the subscript.
Left side: H = 2 ; O = 2
Right side: H = 2 x 2 = 4 ; O = 1 x 2 = 2
Now, notice that the number of O is now balance (both are 2) but the number of H is not (since
You also multiply the left side H by 2. Hence,
Left side: H = 2 x 2 = 4 ; O = 2
Right side: H = 2 x 2 = 4 ; O = 1 x 2 = 2
The equation is now balanced.
Two radioactive nuclides X and Y both decay to stable products. The half-life of X is about a day, while that of Y is about a week. Suppose a radioactive sample consists of a mixture of these two nuclides. If the mixture is such that the activities arising from X and Y are initially equal, then a few days later the activity of the sample will be due ______________.
Answer:Then a few days later the activity of the sample will be due to have more of Nuclides Y than X
Explanation:
This is because half life of nuclide X is about a day which is less than Y having half life of about a week, After a few days, we would observe that X would have disintegrated more while Y will still be predominant since it disintegrate slower than X. The time it takes for X to disintegrate will always be faster than Y.
2. A 47.7 g sample of SrCl2 is dissolved in 112,500.0 g of solvent. Calculate the molality of the
solution.
A. 0.301 m
B. 2.67 m
C. 0.0339 m
D. 3.99 m
Answer:
I have the same question
Explanation:
A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a specific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K?
A. 8.46 kJ
B. 16.1 kJ
C. 27.7 kJ
D. 47.8 kJ
Answer:
27.7 KJ
Explanation:
Q= mC dT
m= 1900 g+0.580 g= 1900.58 g
Q= (1900.58 g * 3.21 KJ/ gK* 4.542 K)
Q=27710 J= 27.7 KJ
Answer:
C) 27.7 kJ
Explanation:
A galvanic cell is powered by the following redox reaction: br2 (l) 2NO(g) (l)(aq) (aq) (aq) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to decimal places.
Final answer:
In the galvanic cell, the reduction of Br₂ to 2Br⁻ occurs at the cathode, while the oxidation of 2NO to N₂O₄ occurs at the anode. The cathode is the silver electrode and the anode is the copper electrode. The silver electrode is the positive electrode and the copper electrode is the negative electrode.
Explanation:
In the galvanic cell powered by the redox reaction between Br₂ and 2NO, the half-reaction that takes place at the cathode is the reduction of Br₂ to 2Br⁻. The half-reaction that takes place at the anode is the oxidation of 2NO to N₂O₄.
For the electrode assignment, the cathode is the electrode where reduction occurs, so it is the silver electrode. The anode is the electrode where oxidation occurs, so it is the copper electrode.
The positive electrode in this galvanic cell is the cathode, which is the silver electrode, and the negative electrode is the anode, which is the copper electrode.
HURRY PLS HELP PLEASE: A 40.0-L sample of fluorine is heated from 363 Kelvin to 459 K. What volume will the sample occupy at the higher temperature?
19.3 L
31.6 L
50.5 L
82.6 L
Final answer:
The volume of a 40.0-L sample of fluorine gas heated from 363 K to 459 K can be found using Charles's Law. After setting up the equation from Charles's Law and solving for the new volume (V2), we find that the volume at the higher temperature is 50.5 L.
Explanation:
To determine the new volume of fluorine gas when heated from 363 K to 459 K, we can use Charles's Law, which states that for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature. Specifically, the formula is V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature.
In this case, the initial volume (V1) is 40.0 L and the initial temperature (T1) is 363 K. The final temperature (T2) is 459 K. Plugging these values into the formula, we have:
40.0 L / 363 K = V2 / 459 K
Multiplying both sides by 459 K to solve for V2 gives us:
V2 = (40.0 L x 459 K) / 363 K
Upon calculation, V2 = 50.5 L. Therefore, the new volume occupied by the sample at 459 K is 50.5 liters.
is 8 inches enough to satisfy a girl
Answer:
Yes
Explanation:
Calculate the number of moles of O2 gas held in a sealed 2.00 L tank at 3.50 atm and 25 ℃.
Answer:
[tex]n=0.286mol[/tex]
Explanation:
Hello,
In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:
[tex]PV=nRT\\\\n=\frac{PV}{RT}[/tex]
Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:
[tex]n=\frac{3.50atm*2.00L}{0.082\frac{atm*L}{mol*K}(25+273)K} \\\\n=0.286mol[/tex]
Best regards.
Write the chemical reaction for nitrous acid in water, whose equilibrium constant is K a . Include the physical states for each species. K a reaction: Write the chemical reaction for the nitrite ion in water, whose equilibrium constant is K b . Include the physical states for each species.
Answer:
HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
Explanation:
According to Brönsted-Lowry acid-base theory, nitrous acid is an acid because it transfers an H⁺ to another compound. The corresponding reaction is:
HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
According to Brönsted-Lowry acid-base theory, nitrite ion is a base because it accepts an H⁺ from another compound. The corresponding reaction is:
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
The chemical reaction for nitrous acid in water is HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
And, the chemical reaction for the nitrite ion in water, is
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
Brönsted-Lowry acid-base theory:
Since nitrous acid should be an acid due to this it transferred an H⁺ to another compound so here the reaction should be HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
Here nitrite ion should have a base due to this, it accepts an H⁺ from another compound. So here the reaction is NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
Learn more about reaction here: https://brainly.com/question/25673311
To how much water should 50mL of 12M hydrochloric acid be added to produce a 4M solution?
Answer:
50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.
Explanation:
According to law of dilution-
[tex]C_{1}V_{1}=C_{2}V_{2}[/tex]
where [tex]C_{1},C_{2},V_{1}[/tex] and [tex]V_{2}[/tex] are initial concentration, final concentration, initial volume and final volume respectively.
Here, [tex]C_{1}=12M[/tex] , [tex]C_{2}=4M[/tex] , [tex]V_{1}=50mL[/tex]
So, [tex]V_{2}=\frac{C_{1}V_{1}}{C_{2}}[/tex] = [tex]\frac{(12M)\times (50mL)}{(4M)}[/tex] = 150 mL
Hence, final volume of HCl solution should be 150 mL.
So, volume of water needed = (150-50) mL = 100 mL
Therefore 50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.
You need to add [tex]\( {150} \)[/tex] mL of water to 50 mL of 12 M hydrochloric acid to produce a 4 M solution.
To determine how much water should be added to 50 mL of 12 M hydrochloric acid (HCl) to produce a 4 M solution, we can use the dilution formula which states:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] and [tex]\( V_1 \)[/tex] are the concentration and volume of the initial solution (12 M and 50 mL, respectively),
- [tex]\( C_2 \)[/tex] and [tex]\( V_2 \)[/tex] are the concentration and volume of the final solution (4 M and the unknown volume of water, respectively).
Let's solve for [tex]\( V_2 \)[/tex]:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
Substitute the given values:
[tex]\[ 12 \times 50 = 4 \times V_2 \][/tex]
Solve for [tex]\( V_2 \)[/tex]:
[tex]\[ 600 = 4 \times V_2 \][/tex]
[tex]\[ V_2 = \frac{600}{4} \][/tex]
[tex]\[ V_2 = 150 \text{ mL} \][/tex]
How many molecules are there in 0.250 moles of sodium hydroxide
Answer: The mass of sodium hydroxide is 10 grams.
Explanation: the atomic weight: 39.997 g/mol
Which of the following is a component of cell theory that is supported by the diagram
Option 4. All living organisms are composed of cells. The component of cell theory that is supported by the diagram is option 4.
The cell theory, which is a fundamental concept in biology, states the following:
All living organisms are composed of one or more cells: This means that cells are the basic structural and functional units of all living things.
The cell is the basic unit of life: Cells are the smallest units that can carry out all the processes necessary for life, including metabolism and reproduction.
All cells arise from pre-existing cells: New cells are produced by the division of existing cells, and no new cells are spontaneously generated.
From the diagram, the first image is a simple unicellular organism and the others are multicellular organism tissues. These shows that all living things are made up of cell/cells depending on the complexity of the organism.
Complete question
What is the molarity of a hydrochloric acid solution, HCl (aq), if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.200 M magnesium hydroxide solution, Mg(OH)2 (aq)? Be sure to write out the balanced chemical reaction between these two compounds.
Answer:
0.333 M
Explanation:
Step 1: Write the balanced equation
2 HCl(aq) + Mg(OH)₂(aq) = MgCl₂(aq) + 2 H₂O(l)
Step 2: Calculate the reacting moles of Mg(OH)₂
25.00 mL of a 0.200 M magnesium hydroxide react. The reacting moles of Mg(OH)₂ are:
[tex]25.00 \times 10^{-3} L \times \frac{0.200mol}{L} = 5.00 \times 10^{-3} mol[/tex]
Step 3: Calculate the reacting moles of HCl
The molar ratio of HCl(aq) to Mg(OH)₂ is 2:1. The reacting moles of HCl are:
[tex]5.00 \times 10^{-3} molMg(OH)_2 \times \frac{2mol}{1molMg(OH)_2} =1.00 \times 10^{-2}molHCl[/tex]
Step 4: Calculate the molarity of HCl
[tex]\frac{1.00 \times 10^{-2}mol}{30.00 \times 10^{-3}L} = 0.333 M[/tex]
a student prepares a dilute solution of sodium hydroxidem, NaOH (aq), starting with 6 M sodium hydroxide. She then titrates a 1.372 g samle of KHP with the dilute sodium hydroxide solution, NaOH (aq), calculate the molar concentration of the sodium hydroxide solution, NAoH (aq)
Answer:
M = 0.3077 M
Explanation:
As I said in the comments, you are missing the required volume of the base to react with the KHP. I found this on another site, and the volume it used was 21.84 mL.
Now, KHP is a compound often used to standarize NaOH or KOH solutions. This is because it contains a mole ratio of 1:1 with the base, so it's pretty easy to use and standarize any base.
Now, as we are using an acid base titration, the general expression to use when a acid base titration reach the equivalence point would be:
n₁ = n₂ (1)
This, of course, if the mole ratio is 1:1. In the case of KHP and NaOH it is.
Now, we also know that moles can be expressed like this:
n = M * V (2)
And according to this, we are given the volume of base and the required mass of KHP. So, if we want to know the concentration of the base, we need to get the moles of the KHP, because in the equivalence point, these moles are the same moles of base.
The reported molar mass of KHP is 204.22 g/mol, so the moles are:
n = 1.372 / 204.22 = 6.72x10⁻³ moles
Now, we will use expression (2) to get the concentration of the diluted base:
n = M * V
M = n / V
M = 6.72x10⁻³ / 0.02184
M NaOH = 0.3077 M
This is the concentration of the dilute solution of NaOH
The molar concentration of sodium hydroxide ( NaOH ) is ; 0.3077 M
Given data :
Volume of base = 21.84 mL = 0.02184 L ( missing data )
Molar mass of KHP = 204.22 g/mol
mass of KHP = 1.372 g
At equivalence point
n1 = n2 ( ratio of KHP to NaOH )
note : n = M * V ---- ( 1 )
First step : calculate the number of moles of KHP
n = mass / molar mass
n = 1.372 / 204.22 = 6.72 * 10⁻³ moles
Determine the molar concentration of NaOH
From equation ( 2 )
M ( molar concentration ) = n / V
= 6.72 * 10⁻³ moles / 0.02184 L
= 0.3077 M
Hence we can conclude that The molar concentration of sodium hydroxide ( NaOH ) is 0.3077 M
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4) What is the difference between tetrahedral bent and trigonal planar bent?
Answer:
A tetrahedral bent molecular geometry involves four electron pairs including two lone pairs and two bonding groups while a trigonal planar bent molecular geometry involves three electron pairs two bonding groups and one lone pair.
Explanation:
According to valence shell electron pair repulsion theory, the number of electron pairs on the central atom of a molecule influences its shape. It follows that, the shape of a molecule is a consequence of arrangement of valence shell electron pairs. Electron pairs must be positioned as far apart in space as possible.
Electron pairs may be bonding pairs or lone pairs. The repulsion between two lone pairs is greater than the repulsion between a lone pair and a bond pair which in turn is greater than the repulsion between two bond pairs. Hence the presence of lone pairs causes much distortion of the expected molecular geometry of the molecule as predicted by VSEPR theory. This distortion usually leads to the assumption of a bent geometry, depending on the expected geometry of the molecule based on VSEPR theory.
A tetrahedral bent molecular geometry involves four electron pairs including two lone pairs and two bonding groups while a trigonal planar bent molecular geometry involves three electron pairs; two bonding groups and one lone pair. For a bent tetrahedral geometry, the bond angle is much less than 109° while for a bent trigonal planar geometry, the bond angle is much less than 120°.
The difference between tetrahedral bent and trigonal planar bent molecular geometries lies in the number of electron pairs around the central atom and the resulting shape.
In a tetrahedral bent geometry, the central atom is surrounded by four electron pairs, which are arranged around the atom in a tetrahedral shape. This arrangement leads to bond angles of approximately 109.5 degrees.
An example of a molecule with a tetrahedral bent shape is water (H2O), where the central oxygen atom has two bonded pairs and two lone pairs of electrons. The lone pairs exert greater repulsion than the bonded pairs, causing the bond angles to be slightly less than the ideal tetrahedral angle, resulting in a bent shape.
On the other hand, in a trigonal planar bent geometry, the central atom is surrounded by three electron pairs, which are arranged in a plane around the atom in a trigonal planar shape.
This arrangement leads to bond angles of approximately 120 degrees. However, when one or more of these electron pairs are lone pairs, the geometry can become bent due to the lone pair-bond pair repulsion being greater than bond pair-bond pair repulsion.
An example of a molecule with a trigonal planar bent shape is sulfur dioxide (SO2), where the central sulfur atom has two bonded pairs and one lone pair of electrons. The presence of the lone pair causes the molecule to have a bent shape with bond angles less than 120 degrees.
In summary, the key differences are:
- Tetrahedral bent involves four electron pairs with a tetrahedral arrangement, while trigonal planar bent involves three electron pairs with a trigonal planar arrangement.
- Tetrahedral bent typically results from two bonded pairs and two lone pairs, leading to a bond angle of slightly less than 109.5 degrees.
- Trigonal planar bent typically results from two bonded pairs and one lone pair, leading to a bond angle of less than 120 degrees.
Therefore, the main difference is the number of electron pairs and the resulting molecular shape and bond angles.
Electrophilic addition to an alkene proceeds via Markovnikov regiochemistry due to the formation of the more stable carbocation intermediate. In the case of conjugated dienes, that is dienes that are separated by one sigma bond, the carbocation that is formed is stabilized additionally by resonance. Addition of the nucleophile to the carbocation intermediate can therefore give two types of products: direct addition to the double bond, also called 1,2-addition, and conjugate addition to the resonance stabilized carbocation, also called 1,4-addition. Allylic carbocation stability is affected by both the nature of the carbocation (primary allylic, secondary allylic, or tertiary allylic) and by the degree of substitution of the double bond. The latter is typically the dominant effect and so a primary allylic carbocation with a trisubstituted double bond is more stable than a tertiary allylic carbocation with a monosubstituted double bond. Electrophilic addition to a conjugated diene is temperature dependent where reaction at or below room temperature typically leads to a mixture of products in which the 1,2 adduct (or direct addition product) predominates, this is termed kinetic control. At elevated temperatures the reactions have time to come to equilibrium and typically the 1,4 adduct (or conjugate addition product) will predominate, this is termed thermodynamic control. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the step by step explanation to the question above.
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimeter containing water initially at 22.3°C. The final temperature of the water is 24.2°C. Ignoring significant figures, calculate the mass of water in the calorimeter.
Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
To calculate the mass of water in the calorimeter, you equalize the heat gained by the water to the heat lost by the heated aluminum (since energy is conserved). You then rearrange the resulting equation to solve for water mass, substituting the known values into the equation.
Explanation:This question is related to the concept of heat transfer in Physics. Here, we are dealing with a piece of aluminium that was heated and then dropped into a calorimeter with water. We want to find the mass of the water contained in the calorimeter.
The heat lost by aluminum will be the heat gained by the water, so we should equalize the heat gained and lost: (mass of Aluminum)*aluminum heat capacity*(Initial temperature-Final temperature) = (water mass)*water heat capacity*(final temperature-initial temperature).
To find the mass of the water, rearrange the equation to solve for it: Mass_water = (mass of Aluminum * aluminum heat capacity * (Initial temperature - Final temperature)) / (water heat capacity * (final temperature - initial temperature)). Plugging in the known values and working through the math should yield the mass of the water in the calorimeter.
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Consider this chemical reaction, where moving from left to right represents moving forward in time. A five panel comic strip. In the first panel, there are ten large red spheres. In the second panel, there are 8 large red spheres and two small blue spheres. In the third panel, there are six large red spheres and four small blue spheres. In the fourth panel, there are four large red spheres and six small blue spheres. In the fifth panel, there are four large red spheres and six small blue spheres. At what point does the reaction first reach equilibrium
Answer:
The reaction reaches equilibrium at the fourth panel.
Explanation:
Chemical Equilibrium is achieved when the overall properties the system seem to be constant, that is, stop changing.
Although, for chemical equilibrium, the right term for this equilibrium is dynamic equilibrium; the rate of forward reaction balances the rate of backward reaction, but concentrations can keep changing.
The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific colour and number of spheres start to become unchanged.
And from the description given in the question,
- In the first panel, there are ten large red spheres.
- In the second panel, there are 8 large red spheres and two small blue spheres.
- In the third panel, there are six large red spheres and four small blue spheres.
- In the fourth panel, there are four large red spheres and six small blue spheres.
- In the fifth panel, there are four large red spheres and six small blue spheres.
It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.
Hence, equilibrium is first reached at the fourth panel.
Hope this Helps!!!
The reaction reaches equilibrium at the fourth panel. A reaction is at equilibrium when the amounts of reactants or products no longer change. Chemical equilibrium is a dynamic process
What is Chemical Equilibrium?It is the rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.
The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific color and number of spheres start to become unchanged.
As per the descriptions of panel given in question:
It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.
Hence, equilibrium is first reached at the fourth panel.
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Determine the pH of a 0.188 M NH 3 solution at 25°C. The K b of NH 3 is 1.76 × 10 -5.
Answer:
pH is 8.52
Explanation:
According to the working in the photo
Acid indigestion is sometimes neutralized with an antacid such as magnesium hydroxide (Mg(OH)2). What products will be released when the antacid is mixed with the hydrochloric acid in the stomach
Answer:
Magnesium chloride and water
Explanation:
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
magnesium chloride water
Consider benzene in the gas phase, C6H6(g)Use the heat of formation, ΔH∘f , values below to answer the questions.
Substance H(g) C(g) C6H6(g) ΔH∘f (kJ/mol ) 217.94 718.4 82.9
Part A
What is the standard enthalpy change for the reaction that represents breaking all the bonds in gaseous benzene, C6H6(g) ?
Express your answer to one decimal place and include the appropriate units.
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Correct
Part B
What is the chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds in gaseous benzene, C6H6(g) ? Indicate the phase of each species in your answer.
Express your answer as a chemical equation including phases.
Answer: 5535.1KJ/mol
Explanation:The equation for breaking of all the bonds in the benzene is written as follows:
C_6 H_6(g) --> 6C(g) +6H (g)
∆H_rxn= ∑moles of product X∆H_PRODUCT -∑moles of reactants x ∆H_REACTANT
={6mols X ∆H_°f(C) + 6mols X ∆H_°f(H)} –{1 mol ∆H_°fC_6 H_6 (g)
={6 X 217.74+ 6X 718.4} – {82.9}KJ/mol
= 5618.04 – 82.9 KK/mol
=5535.1KJ/mol
b) The equation for the breaking of the carbon-carbon bonds in benzene is illustrated below as
C_6 H_6(g) --> 6C---H (g)
A) The standard enthalpy change for the reaction is ; 5535.1 KJ/mol
B) The chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds is : C₆H₆(g) --> 6C---H(g)
A) The standard enthalpy change for the reaction which represents breaking of all bonds
Reaction equation ; C₆H₆(g) --> 6C(g) + 6H (g)
∆H = ∑ ( moles of product * ∆Hpro ) - ( moles of reactant * ∆H reactant )
= ( 6 * 217.94 + 6 * 718.4 ) – ( 82.9) KJ/mol
= ( 5618.04 ) - ( 82.9 ) = 5535.1 KJ/mol
B) The chemical equation for the reaction that corresponds to breaking the carbon-carbon bonds
chemical equation = C₆H₆(g) --> 6C---H(g)
Hence we can conclude that The answers to your questions are 5535.1 KJ/mol and C₆H₆(g) --> 6C---H(g)
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What is the total number of distinct 13C NMR signals that may be observed for the product, methyl-3-nitrobenzoate, and for the reactant, methylbenzoate?What is the total number of distinct 13C NMR signals that may be observed for the product, methyl-3-nitrobenzoate, and for the reactant, methylbenzoate?
Answer:
See explaination
Explanation:
methyl-3-nitrobenzoate = 3-NO2-C6H4-COOCH3
Singlet at 4 ppm = CH3 from ester (COOCH3)
Triplet at 7.6 ppm = Aromatic, H5 proton
Doublet at 8.2 ppm - Aromatic, H4 and H6 protons
Singlet at 8.8 ppm - Aromatic, H2 proton
See attached file for diagrammatic representation and further solution.
Methylbenzoate has 8 distinct carbon environments, therefore it would show 8 distinct 13C NMR signals. Methyl-3-nitrobenzoate, due to its symmetrical structure, has 7 distinct carbon environments, therefore it would show 7 distinct 13C NMR signals.
Explanation:
The product, methyl-3-nitrobenzoate, and the reactant, methylbenzoate, are both organic compounds, and their 13C NMR signals can be determined by the number of carbon environments they have.
Methylbenzoate has 8 distinct carbon environments: benzoate carbon framework (7) + methyl group (1). Hence, in the 13C NMR spectroscopy, methylbenzoate would show 8 distinct signals.
Methyl-3-nitrobenzoate on the other hand, due to the introduction of the nitro group, creates symmetry which changes the carbon environments, thus, it has 7 distinct carbon environments: nitrobenzoate carbon framework (6) + methyl group (1). So, methyl-3-nitrobenzoate would show 7 distinct signals in its 13C NMR spectroscopy.
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g Solid sulfur hexafluoride evaporates when heated at atmospheric pressure rather than liquefying. What is the correct name of the type of phase transition that is represented by this process? a) This process is an example of vaporization. b) This process is an example of deposition. c) This process is an example of sublimation. d) This process is an example of freezing. e) This process is an example of condensation.
Answer: c) This process is an example of sublimation.
Explanation:
When liquid is further heated, the molecules gain more kinetic energy and molecules move farther and convert to gaseous state and the process is called vaporization.
Sublimation is a process of converting a substance from solid state to gaseous state without the formation of liquid at constant temperature.
Deposition is a process of converting a substance from gaseous state to solid state without the formation of liquid at constant temperature.
In liquid the particles are loosely packed as the inter molecular forces of attraction are not that strong. As they are cooled, the kinetic energy of the molecules decreases and thus the molecules start getting closer and convert to solid state. The process is called freezing.