Copper(II) chloride and lead(II) nitrate react in aqueous solutions by double replacement. Write the balanced chemical equation, the overall ionic equation, and the net ionic equation for this reaction. If 11.95 g of copper(II) chloride react, what is the maximum amount of precipitate that could be formed?

Answer:

B. The starting temperature of the water in each cup  

Explanation:

Cindy is trying to see if foam or ceramic is a better insulator. Those are her independent variables.

The other variables , like the starting temperature of the water in each cup, must be controlled variables. If she uses different starting temperatures in each cup, she won't know if it was the temperature or the materials that caused her results.

A. is wrong. The thermometers should be identical but, if they aren't, it will make little difference in the results.

C. is wrong. The ending temperature is room temperature, so it is automatically the same for each cup.

D. is wrong. She is trying to measure the effect of different materials.

Answer:the answer is B(the starting temperature of the water in each cup)

Explanation:

Answer:

The final volume is 990.8 L

Explanation:

Let calculate the moles of gas in the first situation:

P . V = n . R . T

1.20 atm . 2.90 L = n . 0.082 . 293K

(1.20 atm . 2.90 L) / (0.082 . 293K) = 0.145 moles

This are the same moles in the second situation:

P . V = n . R . T

0.003atm . V = 0.145 moles . 0.082 . 250K

V = (0.145 moles . 0.082 . 250K) / 0.003atm

V = 990.8 L

Answer:  The final volume of the balloon is 990 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 1.20 atm

[tex]P_2[/tex] = final pressure of gas = [tex]3.00\times 10^{-3}atm[/tex]

[tex]V_1[/tex] = initial volume of gas = 2.90 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]20^oC=273+20=293K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-23^oC=273-23=250K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{1.20\times 2.90}{293}=\frac{3.00\times 10^{-3}\times V_2}{250K}[/tex]

[tex]V_2=990L[/tex]

Answer:

Scientists have been debating over light being a wave or particle since its recognition.

Sir Issac Newton discovered that light had frequency and other properties. Newton described light to be a particle because it created shadows which were sharp and very clear.

Francesco Maria Grimaldi, claimed that light was a wave. This was because this scientist observed the diffraction of light and hence, claimed light to be a type of wave.

The speed of light is 299 792 458 m / s. Nothing can travel faster than light.

Answer and Explanation:

In order to predict the feasibility of redox processes, standard electrode potentials are majorly employed. Generally, if the electrode potential for the reaction is positive, it is considered to be feasible. However, some conditions affect this statement

The value of E° talks about the feasibility of the reaction under standard conditions only and says nothing about the reaction rate.

A positive value of E° means, the equilibrium constant K is greater than 1; while a negative value of E° means, that it is less than 1.

The attachment below shows the simple analysis of the feasibility of two different reactions A and B, at standard and non standard conditions respectively.

NOTE: Standard conditions for Redox reaction: 298.15K(Temperature), 1 atm(Pressure), 1.0M(Concentration) for both anode and cathode.

Non standard conditions for Redox reaction: Any of the 3 conditions above are changed, especially the concentration.

Answer:

Estimated Average Requirement (EAR).

Explanation:

Hello,

In this case, it is important to consider that Dietary Reference Intakes (DRIs) are reference values to quantitatively estimate the nutrient necessities to be taken for planning and assessing diets for healthy people. On the other hand, the Recommended Dietary Allowance (RDA) is the average daily dietary intake level that is enough to know the nutrient necessity of nearly all (about 98%) healthy individuals in a particular population. The answer is Estimated Average Requirement (EAR) which is a nutrient intake value that is considered to meet the necessity of half (50%) the healthy individuals in a particular population.

Best regards.

Answer:

Presence of lone pairs of electrons

Explanation:

According to VSEPR theory, the presence of lone pairs caused increased repulsion of electron pairs on the valence shell of the centeral atom of the molecule. This decreases or distorts the bond angle. The decrease in bond ange depends on the number of lone pairs present on the valence shell of the central atom of the molecule. Ammonia has only one lone pair hence the bond angle is 107°, water has two lone pairs and the bond angle is 104°. Compare this this with the bond angle of 109° in methane which has only bond pairs and no lone pairs.

The decrease in bond angles from CH4 to NH3 to H2O is due to increasing number of lone pairs on the central atom which cause greater repulsion and smaller bond angles, as per the VSEPR theory.

According to the Valence shell electron-pair repulsion (VSEPR) theory, the progressive decrease in the bond angles in the series of molecules CH4, NH3, and H2O is best accounted for by the presence and arrangement of lone electron pairs on the central atom. CH4 has no lone pairs on the central carbon atom and has a tetrahedral shape with bond angles of 109.5°. In NH3, the nitrogen atom has one lone pair, which decreases the bond angle to less than 109.5°. In H2O, the oxygen atom has two lone pairs, leading to an even smaller bond angle of 104.5°. The VSEPR theory predicts that the electron pairs will arrange themselves to minimize repulsion, and the lone pair-lone pair repulsion is greatest, followed by lone pair-bonding pair, and finally, bonding pair-bonding pair repulsion is the least.

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The scientists use one of their five senses to gather information, they are making an observation.

The active gathering of data from a primary source is observation. Observation of living things makes use of the senses. Using scientific tools to perceive and record data is another way that observation may be used in science.

A technique for gathering information through observation of individuals, situations, or physical qualities as they are occurring naturally.

Using one or more senses, our observational abilities provide us with knowledge about things, occasions, attitudes, and occurrences. Being able to see and learn about the outside world is crucial because it forms the basis of effective communication.

Thus, option A is correct.

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In scientific endeavors, when one of the five senses is used to gather information, it constitutes making an observation. Observations can involve both qualitative and quantitative data, and they provide evidence that scientists use to form conclusions.

When scientists use one of their five senses to gather information, they are A. making an observation. Scientists usually perform observations by using one or more of their five senses to gain an understanding of the properties or behaviors of a substance or a system. These data can be qualitative (descriptive) or quantitative (consisting of numbers). For instance, seeing a plant growing (observation) and measuring its growth over weeks (data collection). From these observations, the scientist can infer conclusions based on the evidence collected. In brain studies, the activity in the brain during specific tasks is observed, which could be the basis for conclusions about brain functionality.

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Answer: If all of the carbon atoms are linked by single covalent bonds and there are no branches, the compounds are called homologous series.

Explanation:

A series of carbon atoms which include different number of carbon atoms but have same functional group are known as homologous series.

Generally, these type of series have a chemical formula as [tex]C_{n}H_{2n+2}[/tex].

No branches are present in this type of series.

For example, [tex]CH_{4}[/tex], [tex]C_{2}H_{6}[/tex], [tex]C_{3}H_{8}[/tex] etc are all homologous series.

Thus, we can conclude that if all of the carbon atoms are linked by single covalent bonds and there are no branches, the compounds are called homologous series.

Answer:

Below.

Explanation:

1.

a.  An alpha particle has 2 protons and 2 neutrons ( a helium nucleus).

b. The charge on an alpha particle is 2+.

c. Atomic symbol used is α or He2+.

d. (no equation shown).

Americanum 241 decays to Neptunium 237 with the loss of 1 alpha particle

Am 241  =  Np 237  + Ne2+

- note the  atomic mass is 4 less under the decay.

Answer:

There is 26.58 grams of gold formed

Explanation:

Step 1: Data given

17.6 A of current are passed through a gold solution for 37.0 min

Molar mass of Au = 196.967 g/mol

Step 2: The equation

Au^3+ + 3e- → Au

Step 3: Calculate coulombs

17.6 Coulomb/s * 37.0 min * 60 sec/min = 39072 Coulombs

1 Faraday = 96500 Coulombs

Step 4: Calculate faraday

39072 Coulombs / 96500 Coulombs / Faraday = 0.40489 Faraday

Step 5: Calculate mass of gold formed

For every 3 Faraday of electricity used up , 1 mole Au is formed

0.40489 Faraday * 1 mole Au/ 3 Faraday = 0.13496 mole Au

196.967 g/mol * 0.13496 mol = 26.58 g Au

There is 26.58 grams of gold formed

The mass of gold that is produced is 26.59 g

Using the formula

[tex]m = \frac{Atomic\ mass}{nF}\times It[/tex]

Where m is the mass

n is the number of equivalents

F is the Faraday constant ( F = 96485 C)

I is the current

and t is the time

From the given information

I = 17.6 A

t = 37.0 min = 37.0 × 60

t = 2220 secs

For gold

Atomic mass = 196.97 g/mol

and n = 3

Putting these parameters into the formula, we get

[tex]m = \frac{196.97}{3 \times 96485} \times 17.6 \times 2220[/tex]

[tex]m = \frac{7696011.84}{289455}[/tex]

m = 26.59 g

Hence, the mass of gold that is produced is 26.59 g

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Answer:

The correct option is option e, both A and C

Explanation:

Ethyl butanoate is an ester. Ester undergoes hydrolysis when heated with water containing dilute acids. Ester hydrolysis gives carboxylic acid and alcohol.

Therefore, ethyl butanoate on hydrolysis gives butanoic acid and ethanol.

Amide also undergoes acidic hydrolysis and yields carboxylic acid and amine.

N-methylbutanamide is an amide. It gives butanoic acid and methylamine when undergoes hydrolysis with H2O in the presence of H2SO4.

Hence, the correct option is option e.

Final answer:

Hydrolysis of ethyl butanoate and N-methylbutanamide in acidic conditions with H₂O and H₂SO₄ will yield butanoic acid; hence, option e) both A and C is correct.

Explanation:

The student's question pertains to the identification of compounds which, upon hydrolysis in an acidic condition with H₂O and H₂SO₄, will yield butanoic acid. To solve this, we need to understand the process of ester hydrolysis under acidic conditions.

Ethyl butanoate (a) is an ester formed from butanoic acid and ethanol. When hydrolyzed, it reverts back to butanoic acid and ethanol. Hence, option (a) is correct. Butyl acetate (b), similarly, is an ester of butanoic acid with butanol and would also yield butanoic acid upon hydrolysis.

N-methylbutanamide (c) is an amide, and on hydrolysis, it gives butanoic acid and a methylamine derivative. Therefore, upon heating with water and sulfuric acid, all listed compounds will yield butanoic acid as a hydrolysis product.

The correct answer to the student's question is e) both A and C.

Final answer:

Approximately 20.7mg of the 50mg sample will be left after 25 days.

Explanation:

The half-life of iodine-131 is 8 days. To determine how much of the 50mg sample will be left after 25 days, we can use the formula:

Amount remaining = Initial amount x (0.5)^(time elapsed / half-life)

Plugging in the values:

Amount remaining = 50mg x (0.5)^(25 / 8) = 50mg x 0.4142 = 20.7mg

Therefore, approximately 20.7mg of the 50mg sample will be left after 25 days.

Answer:

a

Explanation:

Group 15 form trihydrides with the non metal atoms like phosphine, ammonia

The question is incomplete, here is a complete question.

Calculate the standard entropy of vaporization of ethanol at its boiling point 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol.

Answer : The standard entropy of vaporization of ethanol is, 115 J/mol.K

Explanation :

Formula used :

[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]

where,

[tex]\Delta S[/tex] = change in entropy

[tex]\Delta H_{vap}[/tex] = change in enthalpy of vaporization = 40.5 kJ/mol

[tex]T_b[/tex] = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]

[tex]\Delta S=\frac{40.5kJ/mol}{352K}[/tex]

[tex]\Delta S=\frac{40.5\times 10^3J/mol}{352K}[/tex]

[tex]\Delta S=115J/mol.K[/tex]

Therefore, the standard entropy of vaporization of ethanol is, 115 J/mol.K

Answer:

5.8 g of carbon dioxide are produced

Explanation:

The Law of Conservation of Mass states that the mass of the reactants must equal the mass of the products in all chemical reactions.

This is the chemical reaction (combustion)

 CH₄     +   2O₂    →    CO₂   +    2H₂O

12.2 g        14 g             x             20g

Mass in reactants = 12.2 g + 14 g = 26.2 g

Mass in products =  x + 20 g

26.2 g = x + 20g

26.2 g - 20g = x

5.8 g = x

Answer:

0,33atm

Explanation:

For the reaction:

NH₄HS(s) ⇌ H₂S(g) + NH₃(g)

kp is defined as:

kp = 0.11 = P(H₂S) P(NH₃) (1)

Where P(H₂S) and P(NH₃) are partial pressures of each compound.

In equilibrium, if in your system the only addition is of NH₄HS(s), the partial pressures and the concentration of each compound are:

NH₄HS: I - x

-Where I is an initial concentration that is not relevant for the problem and x is the NH₄HS that reacts-

H₂S(g): x

NH₃(g): x

Replacing in (1):

0.11 = X×X

0.11 = X²

0.33 = X

That means P(NH₃) is 0.33 atm

I hope it helps!

The partial pressure of NH3(g) is  0.33 atm.

Number of moles of NH4HS = 46.5 g/51 g/mol = 0.91 moles

Given that;

PV =nRT

P = ?

V = 5.0−L

n =  0.91 moles

R = 0.082 atm LK-1mol-1

T =  250°C + 273 = 523 K

Making P the subject of the formula;

P = nRT/V

P = 0.91 moles ×  0.082 atm LK-1mol-1 × 523 K / 5.0−L

P = 7.8 atm

We must now set up the ICE table;

              NH4HS(s) ⇌   H2S(g)     +          NH3(g)

I            7.8 atm              0                           0

C           -x                      +x                          +x

E          7.8 - x               x                            x

We know that;

Kp = pH2S  × pNH3

Note that NH4HS is a pure solid and does not get into the equation

Kp = 0.11

0.11 = x^2

x = √0.11

x = 0.33 atm

Since partial pressure of H2S = partial pressure of NH3 = x

The partial pressure of NH3(g) = 0.33 atm.

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Answer:

[tex]\rho=1.54\ g/cm^3[/tex]

Explanation:

The expression for density is:

[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]

[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]

M is molar mass of Calcium = 40.078 g/mol

For cubic closest packed structure , Z= 4

[tex]\rho[/tex] is the density

Radius = 197 pm = [tex]1.97\times 10^{-8}\ cm[/tex]

Also, for fcc, [tex]Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm[/tex]

Thus,  

[tex]\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3[/tex]

[tex]\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3[/tex]

[tex]\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3[/tex]

[tex]\rho=\frac{160.312}{104.19483}\ g/cm^3[/tex]

[tex]\rho=1.54\ g/cm^3[/tex]

The density of solid calcium can be calculated by determining the density of its unit cell using the face-centered cubic (FCC) structure. The mass and volume of the unit cell can be calculated using the atomic radius and atomic mass of calcium. Dividing the mass by the volume gives the density of solid calcium.

The density of solid calcium can be calculated by determining the density of its unit cell, which is a face-centered cubic (FCC) structure. In an FCC structure, each unit cell contains 4 atoms. The mass of 4 calcium atoms can be calculated using the atomic mass of calcium, and the volume of the unit cell can be calculated using the atomic radius of calcium. Dividing the mass by the volume gives the density of solid calcium.

The atomic radius of calcium is given as 197 pm, which can be converted to cm by multiplying by 10^-10. The volume of the unit cell can be calculated using the formula V = (edge length)^3. The edge length can be calculated using the diagonal of the face, which is 4 times the atomic radius. The mass of 4 calcium atoms can be calculated using the atomic mass of calcium, which is 40.08 g/mol. Dividing the mass by the volume gives the density of solid calcium.

Density of solid calcium = mass of 4 Ca atoms / volume of unit cell

Keywords: density, solid calcium, unit cell, face-centered cubic (FCC) structure, atomic radius, atomic mass

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Answer:

Decreasing the diameter of the channel by removing aa's. Since Na+ is chemically similar to K+, one can assume the difference must be due to the size of the atom. The K+ ion is larger than the Na+ ion, so reducing the diameter of the channel can allow Na+ to enter while preventing K+ entry. This explains clearly and perfectly how reducing the diameter reduces the

transport of K+ in favor of Na+ transport.

Answers. The correct option is A

Explanation:

The Astronomy manufacturing company is liable to Williams injury only if the company was grossly negligent.

Answer:

Molar concentration: 0,0489M

Explanation:

In this titration of Cu²⁺ you add an excess of I⁻ that reacts with Cu²⁺ producing I₂, this I₂ reacts with Na₂S₂O₃. If you know the I₂ that reacts with Na₂S₂O₃ you can know the I⁻ that reacts with Cu²⁺ and, thus, the quantity of Cu²⁺. The reactions are:

2Cu²⁺ + 4I⁻ → 2CuI + I₂

I₂ + 2S₂O₃⁻ → S₄O₆ + 2I⁻

Moles of S₂O₃⁻ are:

0,01305L×0,15M = 1,96x10⁻³ moles of S₂O₃⁻.

Moles of I₂ are:

1,96x10⁻³ moles of S₂O₃⁻× ( 1 mole of I₂ /  2 moles of S₂O₃⁻) = 9,79x10⁻⁴ moles of I₂

Moles of Cu²⁺ are:

9,79x10⁻⁴ moles of I₂×( 2 moles of Cu²⁺ / 4 moles of I₂) = 4,89x10⁻⁴ moles of Cu²⁺

As volume of the solution was 10,0mL = 0,0100L, the molar concentration of the original solution is:

4,89x10⁻⁴ moles of Cu²⁺ / 0,0100L = 0,0489M

The unknown concentration of the Cu2+ solution can be found by determining the moles of Na2S2O3 at the equivalence point, using this to calculate the moles of Cu2+ from stoichiometry, and then dividing by the volume of the Cu2+ solution in liters.

In this question, a Cu2+ solution, of unknown concentration, is titrated with 0.15 M Na2S2O3. The Cu2+ solution's concentration can be calculated using the data provided. The titration of this Cu2+ solution is complete, or the equivalence point is reached, when 13.05 mL of Na2S2O3 is added. The reaction that occurs is 2Na2S2O3 + Cu2+ -> CuS2O3 + 2Na+, and from the stoichiometry of the reaction, we know that two moles of Na2S2O3 react with one mole of Cu2+ ion.

Using the moles of Na2S2O3 that reacted (moles = Molarity x Volume (in liters), so moles = 0.15 M x 13.05 mL/1000), we can find out the moles of Cu2+ that were present in the 10 mL sample. We can then calculate the molarity of the Cu2+ solution by dividing the moles of Cu2+ by the volume of the solution in liters (0.01 L).

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Answer: Christine Herman & L.G Wade Jr., "2010". Organic Chemistry: Reaction of Alkane, 7e, Pearson Education, Radford University, Radford, VA.

Explanation:

This is an edited book. The Harvard reference style was used in the following order:

Authors name

Year of publication

Title

Edition

Publisher

Place of publication.

Note that the title of book should be italicized with capitalization of first word.

Answer:

44.5 g of Na₂O

Explanation:

The reaction is this one:

2NaCl + MgO  →  Na₂O  +  MgCl₂

Moles of NaCl = Mass / Molar mass

84 g / 58.45 g/m = 1.43 moles

Ratio is 2:1, so if we produce 1 mol of Na₂O, from 2 moles of NaCl; If we have 1.43 moles, we 'll produce the half of moles

1.43 / 2 = 0.72 moles

Molar mass Na₂O = 62 g/m

Mol . molar mass =  0.72 m . 62 g/m = 44.5 g

Answer:

There will be 44.5 grams of sodium oxide (Na2O) produced

Explanation:

Step 1: Data given

Mass of Sodium chloride (NaCl) = 84.00 grams

Magnesium oxide = in excess

Molar mass of NaCl = 58.44 g/mol

Molar mass of sod)ium oxide (Na2O = 61.98 g/mol

Step 2: The balanced equation

2NaCl + MgO → Na2O + MgCl2

Step 3: Calculate moles of NaCl

Moles NaCl = Mass / Molar mass

Moles NaCl = 84.00 grams / 58.44 g/mol

Moles NaCl = 1.437 moles

Step 4: Calculate moles of Na2O

The limiting reactant is NaCl.

For 2 moles NaCl consumed, we need 1 mol MgO to produce 1 mol Na2O and 1 mol of MgCl2

For 1.437 moles of NaCl we'll have 1.437/2 = 0.7185 moles of Na2O

Step 5: Calculate mass of Na2O

Mass Na2O = Moles Na2O * Molar mass Na2O

Mass Na2O = 0.7185 moles * 61.98 g/mol

Mass Na2O = 44.53 grams of Na2O

There will be 44.5 grams of sodium oxide (Na2O) produced

Answer:

C

Explanation:

Gallium is in group thirteen with outermost electron configuration ns2 np1. The highest occupied sub-level is np1 having only one electron which is the situation required in the question.

Answer:

C.  Ga

Explanation:

Answer:

C. 0.20 M Mg ion & 0.40 M Cl ion

Explanation:

MgCl₂ is a ionic salt which is dissociated as this

MgCl₂  →  Mg²⁺  +  2Cl⁻

First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M

Molarity . volume = moles.

0.6 mol/l . 0.2l = 0.12 mol

  MgCl₂  →  Mg²⁺  +  2Cl⁻

0.12mol      0.12         0.24

This moles are also in 400mL of water, so the new concentration is

[Mg²⁺] = 0.12 m/0.6L = 0.2M

[Cl⁻] = 0.24 m/0.6L = 0.4M

Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)

The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M

The correct answer to the question is Option C. 0.20 M Mg ion & 0.40 M Cl ion

we'll begin by calculating the molarity of the diluted solution. This can be obtained as follow:

Volume of stock solution (V₁) = 200 mL

Molarity of stock solution (M₁) = 0.60 M

Volume of diluted solution (V₂) = 200 + 400 = 600 mL

The molarity of the diluted solution can be obtained as follow:

0.6 × 200 = M₂ × 600

120 = M₂ × 600

Divide both side by 600

M₂ = 120 / 600

Thus, the molarity of the diluted (i.e resulting) solution of MgCl₂ is 0.2 M

Next, we shall determine the concentration of magnesium ion, Mg²⁺ in the diluted solution. This is illustrated below:

MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)

From the balanced equation above,

1 mole MgCl₂ dissolves to produce 1 mole Mg²⁺.

Therefore,

0.2 M MgCl₂ will also produce 0.2 M Mg²⁺.

Thus, the concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M.

Finally, we shall determine the concentration of the chloride ion, Cl¯ in the resulting solution.

MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)

From the balanced equation above,

1 mole MgCl₂ dissolves to produce 2 moles of Cl¯

Therefore,

0.2 M MgCl₂ will also produce = 2 × 0.2 = 0.4 M Cl¯

Thus, the concentration of chloride ion, Cl¯ in the resulting solution is 0.4 M.

From the calculations made above:

The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M

Option C. 0.20 M Mg ion & 0.40 M Cl ion gives the correct answer to the question.

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Answer:

a. nitrogen is the correct answer.

Explanation:

Final answer:

Among the given options, nitrogen is a macronutrient required in large amounts for plant growth and development, and is part of vital biomolecules such as carbohydrates, proteins, and nucleic acids.

Explanation:

The macronutrient in question can be identified by knowing that macronutrients are elements that organisms need in relatively large amounts compared to micronutrients which are needed in smaller amounts. The list of macronutrients includes nitrogen (N), phosphorus (P), potassium (K), calcium (Ca), magnesium (Mg), and sulfur (S), while micronutrients or trace elements, such as manganese (Mn), iron (Fe), zinc (Zn), and boron (B), are needed in smaller quantities.

Given the options provided (a. nitrogen b. manganese c. zinc d. boron), the correct answer is a. nitrogen because it is one of the primary macronutrients essential for plant growth and is a part of carbohydrates, proteins, and nucleic acids.

Answer:

The correct answer is 15.

Explanation:

The driving test must be passed with more than 200 points. This means that you can only make a maximum of 15 mistakes to pass the test. This is possible as long as no critical mistakes are made, which are enough to fail the test. These critical mistakes are for example driving faster than allowed, too slow, driving distracted, etc.

Have a nice day!

Answer:

28.7 grams of water

Explanation:

Calorimetry problem:

Q = C . m . ΔT

2400 J = 4.18 J/g°C . m . (30°C - 10°C)

2400 J = 4.18 J/g°C . m . 20°C

2400J = 83.6 J/g . m

2400J / 83.6 g/J = m

28.7 g = m

Final answer:

To calculate the amount of energy absorbed by water when the temperature changes, use the formula: q = mass * specific heat capacity * temperature change.

Explanation:

To calculate the amount of energy absorbed by water, we can use the formula: q = mass * specific heat capacity * temperature change. The specific heat capacity of water is approximately 4.184 J/g°C. In this case, the temperature change is from 10.0°C to 30.0°C, which gives us a ΔT of 20.0°C. We need to convert the mass from grams to kilograms by dividing it by 1000.

So, q = (mass / 1000) * 4.184 J/g°C * 20.0°C.

Substituting the given mass of water (in grams) into the equation, we have: q = (2400 / 1000) * 4.184 J/g°C * 20.0°C. Simplifying the expression gives us the amount of energy absorbed by the water in joules.

Answer:

The answer to your question is 15.7 amu

Explanation:

                          Abundance               Mass

Isotope 1                30%                         15

Isotope 2               70%                          16

Average atomic mass = (Abundance isotope 1 x abundance) +

                                       (Abundance isotope 2 x abundance)

Substitution

Average atomic mass =  (0.30 x 15) + (0.70 x 16)

Simplify

Average atomic mass = 4.5 + 11.2

Result

Average atomic mass = 15.7 amu

Answer: surface tension

Explanation: the tiny weight of insect is not strong enough to break the surface tension of water. So when insects stands or move on water, their feets creates something like dimples on the surface of water which then spring back to propel the insect forward thereby preventing them from sinking.

Answer: The volume of NaOH will be, 12.5 mL

Explanation:

To calculate the volume of base, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HF[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=0.025M\\V_1=25.00mL\\n_2=1\\M_2=0.050M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]1\times 0.025M\times 25.00mL=1\times 0.050M\times V_2\\\\V_2=12.5mL[/tex]

Hence, the volume of NaOH will be, 12.5 mL