Which of the following is the last step in performing a titration
1) finding out which pH indicatorworks
2) determining the concentration of an unknown base
3) finding the number of moles of a product produced in a reaction
4) determining the molecular masses of the products in the reaction
Answer: determining the concentration of an unknown base
Explanation: The last step in performing a titration will lead to the determination of the concentration of the unknown base or an unknown quantity.
The other steps like finding out which pH indicator works, finding the number of moles of a product produced in a reaction and determining the molecular masses of the products in the reaction comes in between the procedure.
Calculate the ph of a 0.10 m solution of barium hydroxide, ba(oh)2. express your answer numerically using two decimal places.
equation is
pH=14+log(.2)
What is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.22 kg of h2 and 31.5 kg of n2?
Given:
Mass of H2 = 5.22 kg = 5220 g
Mass of N2 = 31.5 kg = 31500 g
To determine:
Theoretical yield of NH3
Explanation:
The balanced chemical reaction is:
N2 + 3H2 → 2NH3
1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3
# moles of N2 = 31500 g/ 28 g.mol-1 = 1125 moles
# moles of H2 = 5200 g/ 1 g.mol-1 = 5200 moles
Therefore N2 is the limiting reagent
Based on the stoichiometry:
1 mole of N2 forms 2 moles of NH3
Thus, 1125 moles of N2 will yield : 1125 * 2 = 2250 moles of NH3
Mass of NH3 = 2250 moles * 17 g/mole = 38250 g = 38.3 kg
Ans: Theoretical yield of NH3 = 38.3 kg
Answer : The theoretical yield of ammonia is, 29.58 Kg
Solution : Given,
Mass of [tex]H_2[/tex] = 5.22 Kg = 5220 g
Mass of [tex]N_2[/tex]= 31.5 Kg = 31500 g
Molar mass of [tex]H_2[/tex] = 2 g/mole
Molar mass of [tex]N_2[/tex] = 28 g/mole
Molar mass of [tex]NH_3[/tex] = 17 g/mole
First we have to calculate the moles of [tex]H_2[/tex] and [tex]N_2[/tex] .
Moles of [tex]H_2[/tex]= [tex]\frac{\text{ given mass of }H_2}{\text{ molar mass of }H_2}= \frac{5220g}{2g/mole}=2610moles[/tex]
Moles of [tex]N_2[/tex] = [tex]\frac{\text{ given mass of }N_2}{\text{ molar mass of }N_2}= \frac{31500g}{28g/mole}=1125moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be,
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
From the balanced reaction we conclude that
1 mole of [tex]N_2[/tex] react with 3 moles of [tex]H_2[/tex]
1125 moles of [tex]N_2[/tex] react with [tex]3\times 1125=3375[/tex] moles of [tex]H_2[/tex]
That means [tex]H_2[/tex] is a limiting reagent and [tex]N_2[/tex] is an excess reagent.
Now we have to calculate the moles of ammonia.
From the reaction we conclude that,
3 moles of [tex]H_2[/tex] react to give 2 moles of ammonia
2610 moles of [tex]H_2[/tex] react to give [tex]\frac{2}{3}\times 2610=1740[/tex] moles of ammonia
Now we have to calculate the mass of ammonia.
[tex]\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3[/tex]
[tex]\text{Mass of }NH_3=(1740moles)\times (17g/mole)=29580g=\frac{29580}{1000}=29.58Kg[/tex]
Therefore, the theoretical yield of ammonia is, 29.58 Kg
Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 35.9°c.
The root mean square velocity (RMS velocity) is a method to find a single velocity value for the particles. The average velocity of gas particles in a mixture is found using the root mean square velocity formula:
μrms = sqrt (3RT/M)
Where,
R = universal gas constant = 8.3145 (kg·m^2/s^2)/K·mol
T = temperature = 35.9 ˚C = 309.05 K
M = molar mass = 28 * 10^-3 kg / mol
Substituting the given values into the equation:
μrms = sqrt [(3 * 8.3145 (kg·m^2/s^2) /K·mol) * (309.05 K) / (28 * 10^-3 kg / mol))]
μrms = sqrt (275,313.8813 m^2)
μrms = 524.7 m / s (ANSWER)
Given the formula c1v1=c2v2, where c indicates concentration and v indicates volume, which equation represents the correct way to find the concentration of the dilute solution (c2)
The world's oceans contain approximately 7.6 x 1019 moles of water molecules. Based on this, how many water molecules are there in the world's oceans?
Three beakers contain clear, colorless liquids. one beaker contains pure water, another contains salt water, and another contains sugar water. how can you tell which beaker is which? (
Final answer:
To distinguish between pure water, salt water, and sugar water, conduct tests such as adding silver nitrate to detect chloride ions or observe evaporation rates in a sealed chamber. Salt water may form a precipitate upon adding silver nitrate, while pure water has a higher vapor pressure than sugar water.
Explanation:
To distinguish between a beaker of pure water, salt water, and sugar water, we need to conduct simple experiments that reveal the properties of each liquid as they differ in terms of their chemical components and physical properties.
Adding silver nitrate to a clear colorless solution that forms a pale yellow precipitate indicates the presence of chloride ions, suggesting a salt water solution. A beaker containing pure water would not react, while sugar water would also not produce a precipitate.
When considering evaporation in a sealed chamber, the beaker with pure water will have a higher vapor pressure compared to the sugar water due to the presence of sugar molecules at the surface, which inhibit the evaporation rate. Over time, all of the water from the beaker with higher vapor pressure (pure water) will evaporate and condense in the beaker with the sugar solution.
To differentiate between the three beakers, we must observe the reactions mentioned above (if testing chemically) or deduce based on the properties like vapor pressure and the effects they have on evaporation and condensation processes.
Determine the pOH of a solution with the following 8.3*10^-7 M and classify the solution as acidic basic or neutral
Find the molality of the solution if 42 grams of lithium chloride (LiCl) are dissolved in 3.6 kg of water.
0.275 m
11.7 m
0.990 m
0.857 m
Final answer:
To calculate the molality of the solution, divide the number of moles of lithium chloride (0.990 moles) by the mass of the water in kilograms (3.6 kg), resulting in a molality of 0.275 m.
Explanation:
The molality of a solution is calculated by finding the number of moles of solute and dividing by the mass of the solvent in kilograms. To find the molality of the solution with 42 grams of lithium chloride (LiCl) dissolved in 3.6 kg of water, we first convert grams to moles using the molar mass of LiCl.
The molar mass of LiCl is 42.39 g/mol. So, if we divide 42 grams by the molar mass, we get the number of moles of LiCl:
Moles of LiCl = 42 g / 42.39 g/mol = 0.990 moles of LiCl
We then use this number of moles and the mass of the solvent in kilograms to find the molality:
Molality (m) = Moles of solute / Mass of solvent (kg) = 0.990 moles / 3.6 kg = 0.275 m
Therefore, the molality of the solution is 0.275 m.
Calculate the ph of the resulting solution if 28.0 ml of 0.280 m hcl(aq) is added to
A sample of chromium oxide is 76.5% chromium by weight. what is the simplest formula of the oxide?
A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. 1.27 × 103 1.61 1.61 × 10-3 0.622 622
The concentration of the KCl solution is 1.61 molal.
Explanation:The concentration of a solution can be described by its molality (m), which is the number of moles of solute per kilogram of solvent. To find the molality of the KCl solution, we need to calculate the number of moles of KCl and the amount of water in kilograms. Given that 1.43 mol of KCl is dissolved in 889 g of water, we convert the mass of water to kilograms (889 g = 0.889 kg). Now we can calculate the molality:
Molality (m) = moles of KCl / kilograms of water
Molality (m) = 1.43 mol / 0.889 kg = 1.61 molal
The solute dissociates slightly in the solvent. how will the slight dissociation affect the reported
Calculate the molality of a solution formed by dissolving 34.8 g of lii in 500.0 ml of water. calculate the molality of a solution formed by dissolving 34.8 g of lii in 500.0 ml of water. 0.520 m 0.260 m 0.254 m 0.696 m 0.130 m
The molality of solution formed by dissolving 34.8 g of LiI in 500 mL of water is [tex]\boxed{0.520{\text{ m}}}[/tex].
Further Explanation:
Concentration terms are used to determine concentration of various components present in any mixture. Some of these are mentioned below.
1. Molarity (M)
2. Mole fraction (X)
3. Molality (m)
4. Parts per million (ppm)
5. Mass percent ((w/w) %)
6. Volume percent ((v/v) %)
7. Parts per billion (ppb)
Molality is defined as moles of solute that are present per kilograms of solvent. It is represented by m and its unit is mol/kg. The expression for molality of solution is as follows:
[tex]{\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}[/tex]
Given information:
Mass of LiI: 34.8 g
Volume of water: 500.0 mL
To calculate:
Molality of LiI solution
How to proceed:
Step1: Moles of LiI that are present in 34.8 g of LiI has to be determined. This is done with the help of equation (1).
The formula to calculate the moles of LiI is as follows:
[tex]{\text{Moles of LiI}} = \dfrac{{{\text{Mass of LiI}}}}{{{\text{Molar mass of LiI}}}}[/tex] …… (1)
The mass of LiI is 34.8 g.
The molar mass of LiI is 133.841 g/mol.
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Moles of LiI}} &= \frac{{{\text{34}}{\text{.8 g}}}}{{{\text{133}}{\text{.841 g/mol}}}} \\&= 0.26{\text{ mol}} \\\end{aligned}[/tex]
Step 2: Mass of water is to be evaluated.
At standard temperature and pressure conditions, one liter of water can be considered equivalent to 1 kilograms of water.
Therefore mass of water can be calculated as follows:
[tex]\begin{aligned}{\text{Mass of water}} &= \left( {500.0{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)\left( {\frac{{1{\text{ kg}}}}{{1{\text{ L}}}}} \right) \\&= 0.5{\text{ kg}} \\\end{aligned}[/tex]
Step 3: Molality of LiI solution is to be calculated. This is done with the help of equation (2).
The formula to calculate molality of LiI solution is as follows:
[tex]{\text{Molality of LiI solution}} = \dfrac{{{\text{Moles of LiI}}}}{{{\text{Mass of water}}}}[/tex] …… (2)
Substitute 0.26mol for moles of LiI and 0.5 kg for mass of water in equation (2).
[tex]\begin{aligned}{\text{Molality of LiI solution}} &= \frac{{{\text{0}}{\text{.26 mol}}}}{{{\text{0}}{\text{.5 kg}}}} \\&= 0.520{\text{ m}} \\\end{aligned}[/tex]
Hence, molality of given solution comes out to be 0.520 m.
Learn more:
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Concentration terms
Keywords: concentration, concentration terms, solutions, molarity, molality, LiI, moles, mass, molar mass, 0.520 m, 0.5 kg, 0.26 mol.
A newly discovered element has two isotopes. one has an atomic weight of 120.9038 amu with 57.25% abundance. the other has an atomic weight of 122.8831 amu. what is the atomic weight of the element?
the atomic weight of the element : 122.22 amu
Further explanationThe elements in nature have several types of isotopes
Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.
So Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu
So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom
Mass atom X = mass isotope 1 . % + mass isotope 2.%
An atomic mass unit = amu is a relative atomic mass of 1/12 the mass of an atom of carbon-12.
The 'amu' unit has now been replaced with a unit of 'u' only
for example, Carbon has 3 isotopes, namely ₆¹²C, ₆¹³C, and ₆¹⁴C
Element has two isotopes. 120.9038 amu with 57.25% abundance and the other has an atomic weight of 122.8831 amu
Mass atom element X = mass isotope 1 . % + mass isotope 2.%
Mass atom element X = 120.9038 . 57.25% + 122.8831. 42.75%
Mass atom element X = 69.22 + 52.53 = 122.22 amu
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Based on the experimental results, which solvent provided the second best separation of the analyte?
a. 0.5% nacl solution
b. vinegar
c. 0.2% nacl solution
d. all solvents had the same degree of separation
e. distilled water f. 70% isopropyl alcohol
How many atoms are in a single molecule of water H2O
There are 3 atoms in total. Two come from Hydrogen hence the H2. The third comes from the single Oxygen (O). Hope this helps!
Three atoms make up the water molecule (H2O) as a whole.
What is a molecule?According to the context, the term may or may not include ions that meet this requirement. A molecule is a collection of two or more atoms held together by attractive forces known as chemical bonds.
The same atoms can combine in various ratios to create various molecules. For instance, water (H2O) is made up of two hydrogen atoms and one oxygen atom, whereas hydrogen peroxide (H2O2) is made up of two hydrogen atoms and two oxygen atoms.
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Determine the root-mean square speed of CO2 molecules that have an average kinetic energy of 4.2 x10-21 J per molecule.
The root-mean square speed of the CO₂ molecules is 339 m/s.
The given parameters;
average kinetic energy of CO₂ = 4.2 x 10⁻²¹ J per moleculeThe molecular mass of the CO₂ = (12 + 16x2) = 44 g = 0.044 kg
The average mass of the CO₂ molecule is calculated as;
[tex]average \ mass \ = \frac{molecular \ mass}{Avogadro's \ number} = \frac{0.044}{6.02 \times 10^{23}} = 7.31 \times 10^{-26} \ kg[/tex]
The root-mean square speed is calculated by applying kinetic energy equation;
[tex]E = \frac{1}{2} mv^2\\\\v^2 = \frac{2E}{m} \\\\v= \sqrt{\frac{2E}{m}} \\\\v = \sqrt{\frac{2(4.2 \times 10^{-21})}{7.31\times 10^{-26}}}\\\\v = 338.99 \ m/s \ \approx 339 \ m/s[/tex]
Thus, the root-mean square speed is 339 m/s.
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Write a balanced chemical equation for the reaction. feso4 and pb(no3)2 express your answer as a chemical equation. identify all of the phases in your answer.
Answer: The balanced chemical equation is written above.
Explanation:
Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.
When ferrous sulfate reacts with lead (II) nitrate, leads to the production of iron (II) nitrate and lead (II) sulfate.
The chemical equation for the above reaction follows:
[tex]FeSO_4(aq.)+Pb(NO_3)_2(aq.)\rightarrow Fe(NO_3)_2(aq.)+PbSO_4(s)[/tex]
By Stoichiometry of the reaction:
1 mole of aqueous solution of ferrous sulfate reacts with 1 mole of aqueous solution of lead (II) nitrate to produce 1 mole of aqueous solution of iron (II) nitrate and 1 mole of solid lead (II) sulfate.
Hence, the balanced chemical equation is written above.
An old sample of concentrated sulfuric acid to be used in the laboratory is approximately 98.1 percent h2so4 by mass. calculate the molality and molarity of the acid solution. the density of the solution is 1.83 g/ml.
The molality of a 98.1% sulfuric acid solution is calculated to be 526.32 mol/kg and the molarity is found to be 18.32 M.
Explanation:To find the molality and molarity of sulfuric acid in a solution, we need to know the mass and volume of the solution. Given that the solution is 98.1% sulfuric acid by mass, we can say that for every 1000g (or 1kg) of solution, there are 981g of sulfuric acid. The molecular weight of sulfuric acid is 98.08 g/mol, so we can convert this to moles to find the molality.
Molality = moles of solute / mass of solvent in kg
The moles of sulfuric acid = 981g / 98.08 g/mol = 10 mol. The mass of the water in the solution is 1000g - 981g = 19g, or 0.019kg. So, the molality of the solution is 10 mol / 0.019 kg = 526.32 mol/kg.
To find the molarity, we need to know the volume of the solution. Given the density of the solution is 1.83 g/ml, we can say that 1kg of solution is 1000g / 1.83g/ml = 546ml = 0.546L. So, the molarity = moles of solute / volume of solution in liters = 10 mol / 0.546 L = 18.32 M.
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How much heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k?
528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.
What is heat ?Heat is the result of the movement of kinetic energy within a material or an item, or from an energy source to a material or an object. Radiation, conduction, and convection are the three mechanisms through which such energy can be transferred.
Heat is the quantity of energy that is transferred from one structure to its surrounds as a result of a change in temperature. Heat is the transfer of kinetic energy between one media or item and another, or from an energy source to a medium or object.
The total amount of heat required
∑q = q1 + q2 + q3
q1 = mCΔT
= 475 g x 0.902 J/g-oC x ( 933.47 - 295 ) K
= 273,552.47J
= 273 kJ
ΔT = ( 933.47 - 295 ) K
= 638.47 K
ΔT = ( 933.47 K -273) - (295 K - 273)
= 638.47°C
For ΔT temperature in K is same as in °C
q2 = mol ΔHf
mol = 475 g x 1 mol /27g
= 17.6 mol
q2 = mol ΔHf = 17.6 mol x 10.79 kJ/mol
= 189.9 kJ
q3 = mCΔT
= 0.475 kg x 1.18 kJ/kg-oC (1050 - 933.47)K
= 65.3 kJ
∑q = q1 + q2 + q3
= 273 + 189.9 + 65.3
= 528.2 kJ
Thus,528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.
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At the start of a measurement, a radioisotope has 10,000 unstable nuclei. over 2 days 5,000 of these unstable nuclei undergo radioactive decay to stable nuclei. what is the half life of this radioisotope?
Find the molality of the solution if 2.5 moles of CaBr2 are dissolved in 5 kg of water.
-500 m
-5 x 10-4 m
-0.5 m
-2 m
How many grams of iki would it take to obtain a 100 ml solution of 0.300 m iki? how many grams of iki would it take to create a 100 ml solution of 0.600 m iki?
0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.
In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
Final answer:
To obtain a 100 mL solution of 0.300 M KI, it would take approximately 4.98 grams of KI. To create a 100 mL solution of 0.600 M KI, it would take approximately 9.96 grams of KI.
Explanation:
To determine how many grams of KI are needed to create a 100 mL solution of 0.300 M KI, we can use the equation:
Molarity (M) = moles of solute / volume of solution (L)
First, convert the volume from mL to L: 100 mL = 0.100 L
Then, rearrange the equation to solve for moles of solute:
moles of solute = Molarity x volume of solution
moles of solute = 0.300 M x 0.100 L
moles of solute = 0.030 mol KI
Finally, use the molar mass of KI (166 g/mol) to convert moles to grams:
grams of solute = moles of solute x molar mass
grams of solute = 0.030 mol KI x 166 g/mol
grams of solute = 4.98 g KI
Therefore, it would take approximately 4.98 grams of KI to obtain a 100 mL solution of 0.300 M KI.
Similarly, to create a 100 mL solution of 0.600 M KI, the calculation would be:
moles of solute = 0.600 M x 0.100 L
moles of solute = 0.060 mol KI
grams of solute = 0.060 mol KI x 166 g/mol
grams of solute = 9.96 g KI
It would take approximately 9.96 grams of KI to create a 100 mL solution of 0.600 M KI.
If δh°rxn and δs°rxn are both positive values, what drives the spontaneous (favored) reaction and in what direction at standard conditions?
If δh°rxn and δs°rxn are both positive values, the reaction is spontaneous and favored at standard conditions.
Explanation:If δh°rxn and δs°rxn are both positive values, the reaction is spontaneous and favored at standard conditions. This is because a positive value of δh°rxn indicates an exothermic reaction, releasing energy, while a positive value of δs°rxn indicates an increase in disorder or randomness, which is favorable for a reaction to occur. Therefore, both factors - the release of energy and the increase in disorder - drive the reaction in the forward direction.
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Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are added. at equilibrium, 0.100 mol of nh3 is present. calculate the equilibrium concentrations of n2 and h2 for the reaction: 2 nh3(g) n2(g) + 3 h2(g). n2 1.25 mol/l h2 1.50 mol/l find kc for the reaction. kc = 422 m2 (b) in a different 1.00-l container at the same temperature, equilibrium is established with 8.34 ✕ 10−2 mol of nh3, 1.50 mol of n2, and 1.25 mol of h2 present. calculate kc for the reaction: nh3(g) 1 2 n2(g) + 3 2 h2(g). 20.5 m (c) what is the relationship between the kc values in parts (a) and (b)? why aren't these values the same?
In the formation of ammonia, different initial amounts of reactants result in different Kc values at equilibrium. While the temperature is constant, varying concentrations of reactants and products lead to different equilibrium constants Kc.
Explanation:The chemical equation is given by 2NH3(g) = N2(g) + 3H2(g). In the first scenario, you start with 1.30 mol of N2 and 1.65 mol of H2 in a 1.00 L container, with 0.100 mol of NH3 at equilibrium. This implies a decrease in N2 and H2 by x amount and an increase in NH3 by 2x. Given the stoichiometric relationships, we can calculate that the equilibrium concentrations of N2 and H2 are 1.20 M and 1.30 M respectively.
Subsequently, we can calculate Kc using the equilibrium concentrations, Kc = [NH3]^2/(N2][H2]^3) = (0.10)^2/(1.20 * (1.55)^3) = 422 M^-2.
In the second scenario, the same calculations result in a Kc = 20.5 M^-2. The discrepancy in the Kc values arises due to the differences in the initial amounts of reactants and the products formed. Although the temperature is constant and the reaction is the same, the amount of reactants used and the products formed vary which impacts the Kc value.
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Which is a Not scientific question?
What is the pH of a solution of potassium hydroxide?
Which base is the most exciting to study?
Will HF neutralize a solution of HCl?
Does adding sodium chloride change the pH of a solution?
One mole of an ideal gas is sealed in a 22.4-l container at a pressure of 1 atm and a temperature of 273 k. the temperature is then increased to 311 k , but the container does not expand. what will the new pressure be? part a the most appropriate formula for solving this problem includes only which variables? enter the required variables, separated by commas (e.g., p,v,t).
The appropriate formula for solving this problem includes the variables pressure and temperature (P, T).
Explanation:To solve this problem, we can use the combined gas law formula, which includes the variables pressure (P), volume (V), and temperature (T).
The combined gas law formula is as follows: P1V1/T1 = P2V2/T2
In this problem, the volume is constant, so we only need to consider pressure and temperature. Therefore, the appropriate formula for solving this problem includes the variables pressure and temperature (P, T).
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1. After calculation, the new pressure of the gas is approximately 1.139 atm. 2.The variables involved are pressure (P) and temperature (T).
1. To determine the new pressure of the gas after the temperature increase while keeping the volume constant, we should use the combined gas law, which states:
P₁/T₁ = P₂/T₂
Given:
Initial pressure (P₁) = 1 atmInitial temperature (T₁) = 273 KFinal temperature (T₂) = 311 KWe need to find the final pressure (P₂)Rearrange the formula to solve for P₂:
[tex]\[P_2 = P_1 \cdot \frac{T_2}{T_1}\][/tex]
Substitute the given values:
[tex]P_2 = 1 \, \text{atm} \cdot \left( \frac{311 \, \text{K}}{273 \, \text{K}} \right)[/tex]
Calculate:
P₂ ≈ 1.139 atm
Therefore, the new pressure will be approximately 1.139 atm.
2. The most appropriate formula for solving this problem includes only the following variables: P, T.
How many grams are in 4.00 mol of sodium chloride (nacl)?
Mole measure the number of elementary entities of a given substance that are present in a given sample. The mass of 4.00 mole of sodium chloride is 233.76g.
What is mole?The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity of amount of substance. One mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.
Mathematically,
mole of sodium chloride = mass ÷ Molar mass of sodium chloride
number of mole of sodium chloride=4mol
Molar mass sodium chloride =58.44 g/mole
Substituting the given values in the above equation, we get
4mole of sodium chloride=mass mass of sodium chloride÷ 58.44 g/mole
mass of sodium chloride = 4× 58.44 g/mole
mass of sodium chloride= 233.76g
Therefore the mass of 4.00 mole of sodium chloride is 233.76g.
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find the midpoint of AB if A(-3,8) and B(-7,-6)
The midpoint of AB is (-5,1), if A(-3,8) and B(-7,-6).
How to find out the midpoint of coordinates?The midpoint of coordinates is found by measuring the distance between two endpoints and dividing the obtained result by 2. Apart from this, one more method is there which is to add the two X-coordinates of the endpoints and divide them by 2. The same concept is applied to the Y-coordinates as well.
According to the question,
A = (-3,8) and B = (-7,-6).
The formula is as follows:
M(x,y) = ((x₁+x₂)/2 , (y₁+y₂)/2).
Putting the above values in the mentioned formula, we get:
M(x,y) = ((-3-7)/2 , (8-6)/2).
M(x,y) = (-5 , 1).
Therefore, the midpoint of AB is (-5,1), if A(-3,8) and B(-7,-6).
To learn more about the Midpoint of coordinates, refer to the link:
https://brainly.com/question/28308761
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