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Norma-Jean [14]

2 years ago

7

For the reaction below determine the ▲H for the reaction and state whether the reaction was endothermic or exothermic. Show your

work.
C2H5OH + 3 O2 → 2 CO2 + 3H2O

C-C = 83 kcal; C-H = 99 kcal; C-O = 84 kcal; O-H = 111 kcal; C=O 192 kcal; O=O = 119 kcal

I got stuck on this and need help please, thank you so much!

1 answer:

scoundrel [369]2 years ago

6 0

Answer:exothermice

Explanation: it is relaeing heat not keeping it in

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A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

What weight of sodium formate must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50?

tester [92]

Answer:

156,4 g of sodium formate

Explanation:

The pka of the formic acid is 3,74. Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] <em>(1)</em>

Where A⁻ is the conjugate base (Formate) and HA is the formic acid.

4.00L of 1.00M formic acid contain:

4.00L × (1.00mol /L) = 4.00 moles

Replacing these moles, the desired pH and the pka value in (1):

3,50 = 3,74 log₁₀ [A⁻] / 4,00 moles

-0,24 = log₁₀ [A⁻] / 4,00 moles

0,575 = [A⁻] / 4,00 moles

<em>2,30 moles = [A⁻] </em>

That means you need 2,30 moles of formate (Sodium formate), to produce the desired buffer. As the molar mass of sodium formate is 68,01g/mol, the weight of sodium formate that must be added is:

2,30mol×(68,01g/mol) =<em> 156,4 g of sodium formate</em>.

I hope it helps!

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Explain, List the steps that show how the sunprovides the energy for surface ocean currents.

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Equation 3x-5=-2x+ 10?<br> O<br> 0<br> x= 5<br> -5 = x<br> -15 = -5x<br> -5x = 15

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Answer:

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Answer:

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3 years ago

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