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bezimeni [28]
3 years ago
6

if you help me ill give you more brainly's. plzz help.At 1 atm, how much energy is required to heat 37.0 g of H2O(s) at –10.0 °C

to H2O(g) at 155.0 °C? Helpful constants can be found here.
Chemistry
1 answer:
artcher [175]3 years ago
4 0

Answer:

116.3 kJ

Step-by-step explanation:

Three heat transfers are involved

q = Heat to warm ice + heat to melt ice + heat to warm water + heat to evaporate water + heat to warm steam

q =      q₁      +     q₂        +     q₃       +     q₄          +     q₅

q = mC₁ΔT₁ + mΔH_fus + mC₃ΔT₃ + mΔH_vap + mC₅ΔT₅

<em>Step 1</em>: Calculate q₁

m = 37.0 g

C₁ = 2.010 J·°C⁻¹g⁻¹

ΔT₁ = T_f – T_i  

ΔT₁ = 0.0 – (-10.0)

ΔT₁ = 10.0 °C  

q₁ = 37.0 × 2.010 × 10.0  

q₁= 743.7 J

q₁= 0.7437 kJ

===============

<em>Step 2</em>. Calculate q₂

ΔH_fus = 334 J/g

q₂ = 37.0 × 334

q₂ = 12 360 J

q₂ = 12.36 kJ

===============

Step 3: Calculate q₃

C₃ = 4.179 J·°C⁻¹g⁻¹

ΔT₃ = T_f – T_i  

ΔT₃ = 100 – 0  

ΔT₃ = 100 °C

q₃ = 37.0 × 4.179 × 100  

q₃ = 15 460 J

q₃ = 15.46 kJ

===============

<em>Step 4</em>. Calculate q₄

ΔH_vap = 2260 J/g

q₄ = 37.0 × 2260

q₄ = 83 620 J

q₄ = 83.62 kJ

===============

<em>Step 5</em>. Calculate q₅

C¬₅ = 2.010 J·°C⁻¹g⁻¹

ΔT₅ = T_f – T_i  

ΔT₅ = 155.0 – 1000  

ΔT₅ = 55.0 °C

q₅ = 37.0 × 2.010 × 55

q₅ = 4090 J

q₅ = 4.090 kJ

===============

Step 6. Calculate q

q = 0.7437 + 12.36 + 15.46 + 83.62 + 4.090

q = 116.3 kJ

The heat required is 116.3 kJ.

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nadezda [96]

Answer: The amount of water produced is 9.3 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

CH_4+2O_2\rightarrow CO_2+2H_2O

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4 0
2 years ago
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Answer:

1.21x10^{-3} M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x10^{-3}

H = \frac{0.21}{1.38x10^{-3} }

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = 1.21x10^{-3} M

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