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Vikentia [17]
3 years ago
10

Carbon dioxide (CO2) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "s

upercritical fluid" that is a much safer dry-cleaning agent than tetrachloroethylene. At a certain pressure, the density of supercritical CO2 is 0.469 g/cm3. What is the mass of a 25.0-mL sample of supercritical CO2 at this pressure?
Chemistry
1 answer:
Lunna [17]3 years ago
5 0

Answer:

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

Explanation:

Step 1: Given data

The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)

The sample hasa volume of 25.0 mL

Step 2: Calculating mass of the sample

The density is the mass per amount of volume

0.469g/cm³ = 0.469g/ml

The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

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What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze
enyata [817]

Answer : 17.12 g

Explanation:\Delta T =k_b\times m

\Delta T = elevation in boiling point

k_b = boiling point elevation constant

m= molality

molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}

given \Delta T=6.28^{\circ}C

Molar mass of solute = 46.0 gmol^{-1}

Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

molality=\frac{x}{46gmol^{-1}\times0.15kg}

6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}

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3 0
3 years ago
What volume of a 3.0 M stock solution of H2SO4 is needed to prepare 2.8 L of a 1.6 M H2SO4 solution?
Georgia [21]
Use M1V1 = M2V2 to solve
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4 0
3 years ago
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Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to
snow_lady [41]

Answer:

\displaystyle \text{p} K_a \approx 3.856

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:


\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right]  & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

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[AR] 3d10 4s2 is the configuration for zinc

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