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3 years ago

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Chemistry

1 answer:

lesya692 [45]3 years ago

7 0

Answer:

Student 3

Explanation:

This question lets us know something about how the density of a gas varies with temperature.

When a gas is heated, its molecules spread out and expand. When this happens, the volume of the gas increases. Remember that density is defined as mass/volume. Thus as the volume increases, the density of the gas decreases.

Therefore, the carbon dioxide rose up because the heat expanded the gas and it became less dense.

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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

<u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

<u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

<u>Initial:</u> 0.1372 0.04514

<u>Final:</u> 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

Which part of the cell controls many functions of the cell and stores DNA?

kari74 [83]

Answer:

The Nucleus...

it perform many activities and stores DNA in a cell.

6 0

3 years ago

Read 2 more answers

A total solar eclipse occurs because the moon can totally block the sun. If the moon was smaller or farther from earth, a total

Setler79 [48]

Answer:

both

Explanation:

id say that it could occur but also not as much. the moon would be smaller and further from the earth to where we would barely be able to see it. if the full moon is barely visible then im sure the total solar eclipse wouldn't be as noticeable as it is now. but thats just my opinion

6 0

2 years ago

A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio

Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

$nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol$

$nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol$

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

NaOH + HCl ⇒ NaCl + H₂O

Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0

Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³

Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³

The concentration of NaOH is:

$[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M$

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0

3 years ago

A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

Andreas93 [3]

<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>

The reaction between NaOH and H₂SO₄ is a neutralization reaction.
The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>

Number of moles = Mass ÷ Molar mass
Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

= 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>

From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
Thus, Moles of H₂SO₄ = moles of NaOH × 2

= 0.33 moles × 2

= 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>

When given the molarity of an acid and the number of moles we can calculate the volume of the acid.

That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

= 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

7 0

4 years ago