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2 years ago

How much heat is required to heat 9.61g of ethanol (CH3CH2OH) from 24.10C to 67.30C?

Chemistry

1 answer:

Licemer1 [7]2 years ago

8 0

Answer:

a.)

To warm the liquid from 35°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - 35) = 4154 J

To vaporize the liquid at 78°C:

(38.56 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 35.154 kJ

Total:

4.154 kJ + 35.154 kJ = 39.3 kJ

b.)

To warm the solid from -155°C to -114°C:

(0.97 J/g-K) x (42.0 g) x (-114°C - (-155°C)) = 1670 J

To melt the solid at -114°C:

(5.02 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 4.5766 kJ

To warm the liquid from -114°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - (-114)) = 18547 J

To vaporize the liquid at 78°C:

35.154 kJ (as in part a.)

Total:

1.670 kJ + 4.5766 kJ + 18.547 kJ + 35.154 kJ = 59.9 kJ

Explanation:

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2 years ago

Which of these tasks falls under the responsibility of a forensic scientist?

Varvara68 [4.7K]

Answer:

Training officers in how to properly collect evidence

Explanation:

Forensic science is an interesting branch of science that involves the use of scientific procedures to solve a crime case. It encompasses collection of physical evidence from the crime scene and analyzing it in a laboratory using scientific means.

A forensic scientist is the individual in charge of performing these scientific procedures. His/her major role is to run the scientific analysis of the physical evidence brought in by the officers, however, he/she can also perform the task of training officers in how to properly collect evidence, in order not to damage the evidence or render it invalid for use.

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2 years ago

What is the balanced chemical equation for the reaction used to calculate ΔH∘f of SrCO3(s)? If fractional coefficients are requi

Pani-rosa [81]

Answer:

Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)

Explanation:

The standard enthalpy of formation (ΔH°f) is the energy involved in the formation of 1 mole of a substance from its elements in their most stable states. The chemical equation for the formation of SrCO₃(s) is the following.

Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)

6 0

2 years ago

The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T

kifflom [539]

Answer : The equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

First we have to calculate the standard free energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]$

where,

$\Delta G^o$ = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]$

$\Delta G^o=-90.72kJ/mol$

Now we have to calculate the value of $K_p$

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G_^o$ = standard Gibbs free energy = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p$

$K_p=7.98\times 10^{15}$

Now we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium $SO_2$ pressure be, x

Pressure of $SO_2$ = Pressure of $H_2S$ = x

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

$7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}$

$x=3.93\times 10^{-5}torr$

Thus, the equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

4 0

2 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago