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kirza4 [7]
2 years ago
14

To standardize an H2SO4 solution, you put 20.00 mL of it in a flask with a few drops of indicator and put 0.450 M NaOH in a bure

t. The buret reads 0.63 mL at the start and 44.73 mL at the end point. Find the molarity of the H2SO4 solution. Start by writing a balanced chemical equation for this neutralization reaction. Question 5 options: 0.250 M
Chemistry
1 answer:
hram777 [196]2 years ago
7 0

To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.

We want to standardize a H₂SO₄ solution with NaOH. The neutralization reaction is:

H₂SO₄ + 2 NaOH ⇒ Na₂SO₄ + H₂O

The buret, which contains NaOH, reads initially 0.63 mL, and 44.73 mL at the endpoint. The used volume of NaOH is:

V = 44.73 mL - 0.63 mL = 44.10 mL

44.10 mL of 0.450 M NaOH are used for the titration. The reacting moles of NaOH are:

0.04410 L \times \frac{0.450mol}{L} = 0.0198 mol

The molar ratio of H₂SO₄ to NaOH is 1:2. The moles of H₂SO₄ that react with 0.0198 moles of NaOH are:

0.0198 mol NaOH \times \frac{1molH_2SO_4}{2molNaOH} = 0.00990 molH_2SO_4

0.00990 moles of H₂SO₄ are in 20.00 mL of solution. The molarity of H₂SO₄ is:

[H_2SO_4] = \frac{0.00990mol}{0.02000} = 0.495 M

To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.

Learn more: brainly.com/question/2728613

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It turns from a liquid to a gas
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Why do real gases not behave exactly like ideal gases?
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The theory assumes that collisions between gas molecules and the walls of a container are perfectly elastic, gas particles do not have any volume, and there are no repulsive or attractive forces between molecules .

5 0
3 years ago
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A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t
andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)

<u>Where:</u>

<em>μ (l): is the chemical potential of 2-propanol in solution    </em>

<em>μ° (l): is the chemical potential of pure 2-propanol   </em>

<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>

<em>T: is the temperature = 82.3 °C = 355.3 K </em>

<em>x: is the mole fraction of 2-propanol = 0.41 </em>

\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)

\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}

\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}  

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

I hope it helps you!    

8 0
3 years ago
What volume, in milliliters, of calcium chloride 15.0 M stock solution would you use to
Kamila [148]

<u>Answer:</u> The volume of stock solution of calcium chloride required is 10 mL.

<u>Explanation:</u>

A solution consists of solute and solvent. A solute is defined as the component that is present in a smaller proportion while the solvent is defined as the component that is present in a larger proportion.

To calculate the amount of solute needed, the formula used is:

C_1V_1=C_2V_2             ....(1)

where,

C_1\text{ and }V_1 are the concentration and volume of stock solution of calcium chloride

C_2\text{ and }V_2 are the concentration and volume of diluted solution of calcium chloride

Given values:

C_1=15.0M\\C_2=0.300M\\V_2=500mL

Plugging values in equation 1:

15.0\times V_2=0.300\times 500\\\\V_2=10mL

Hence, the volume of stock solution of calcium chloride required is 10 mL.

7 0
2 years ago
I’m so lost please help
ExtremeBDS [4]

C=4

D=-19

Explanation:

2exp-25/5exp-7= 4exp-19

7 0
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