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yawa3891 [41]
3 years ago
7

How much heat will be released when 10.0 g of hydrogen peroxide decomposes according to the following reaction: 2H2O2(l) 2H2O(l)

+ O2(g) ΔH = -190 kJ
A. 26 kJ
B. 22 kJ
C. 28 kJ
D. 24 kJ
Chemistry
1 answer:
kramer3 years ago
5 0
C. 28 KJ

AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2

0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

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What is the molarity of a solution in which 250 grams of NaCl are dissolved in 2.9 L of water?
Maksim231197 [3]

Answer:

About 1.48 M.

Explanation:

The formula for molarity is mol/L.

So firstly, you must find the amount of moles in 250 grams of NaCl.

I do this by using stoichiometry. First, I find how nany grams are in a single mole of NaCl. This is around 58.44 grams/mole. Now that I know this, I can now use a stoich table. (250 g NaCl * 1 mol NaCl / 58.44 g NaCl). I plug this into my calculator.

I get that 250 grams of NaCl is equal to about 4.28 moles.

Now I just plug into the formula!

4.28 moles/2.9 L = about 1.48

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