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Oksi-84 [34.3K]

2 years ago

12

The common name for the compound ch3ch2ch2 och2ch3 is

1 answer:

Diano4ka-milaya [45]2 years ago

4 0

Caesium sulfate (cesium sulfate) is the inorganic compound and salt with the formula Cs2SO4. It is a white water-soluble solid that is used to prepare dense aqueous solutions for use in isopycnic (or "density-gradient") centrifugation. It is isostructural with potassium salt.

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Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul

BlackZzzverrR [31]

Answer:

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Explanation:

Assuming that all caculations are at normal pressure and -1.72°C :

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Where

$n$ is the number of moles of hydrogen

$n$ is the mass of hydrogen

$\rho$ is the density of hydrogen

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3 years ago

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3 years ago

What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C

Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at $26.5^oC$ = ?

$P_2$ = final pressure at $-5.1^oC$ = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $26.5^oC=273+26.5=299.5K$

$T_2$ = final temperature = $-5.1^oC=273+(-5.1)=267.9K$

Now put all the given values in this formula, we get

$\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]$

$\log (\frac{P_1}{100})=-0.576$

$\frac{P_1}{100}=0.265$

$P_1=26.5mmHg$

Thus the vapor pressure of $CS_2CS_2$ in mmHg at 26.5 ∘C is 26.5

7 0

3 years ago