A reaction gives off 35.6 kJ how many calories is this?

Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enth

stealth61 [152]

Answer : The value of $\Delta H_{vap}$ is 28.97 kJ/mol

Explanation :

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

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

where,

$P_1$ = vapor pressure at temperature $-21.0^oC$ = 462.7 mmHg

$P_2$ = vapor pressure at temperature $-44.0^oC$ = 140.5 mmHg

$\Delta H_{vap}$ = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $-21.0^oC=[-21.0+273]K=252K$

$T_2$ = final temperature = $45^oC=[-41.0+273]K=232K$

Putting values in above equation, we get:

$\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol$

Therefore, the value of $\Delta H_{vap}$ is 28.97 kJ/mol

4 0

3 years ago

A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.314 kg of water?

Mariulka [41]

Answer:

g NaCl = 424.623 g

Explanation:

<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent

∴ solvent: H2O

∴ mass H2O = 2.314 Kg

mol NaCl:

⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl

∴ mm NaCl = 58.44 g/mol

⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl

5 0

3 years ago

Question 3 of 10

PIT_PIT [208]

Answer:

the moluculer formula is the answer

Explanation:

4 0

3 years ago

Please help. im freaking out rn. i have like 40 missing assignments please

Katarina [22]

Answer:

I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry

5 0

3 years ago

Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH

Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{12 g}{60 g/mol}$

= 0.2 mol

Molarity of acetic acid is calculated as follows.

Density = $\frac{mass}{volume}$

1 g/ml = $\frac{100 g}{volume}$

volume = 100 ml

Hence, molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.2 mol}{0.1 L}$

= 2 mol/l

As reaction equation for the given reaction is as follows.

$NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O$

So, moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

$x \times 1 M = 10 mL \times 2 M$ (as 1 L = 1000 ml)

x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.

6 0

3 years ago