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frez [133]
3 years ago
15

A cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the system is raise

d by 74 K. Calculate the heat lost or gained by the system. Calculate your answer to three significant figures and express you answer in J. Heat capacity for Helium = 20.8 J/(mol K) Molecular weight for Helium = 4 g/mol
Chemistry
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

There is 96200 J or 96 kJ of heat released.

Explanation:

<u>Step 1: </u>Data given

Mass of helium = 250 grams

Temperature = 200 K

Temperature is raised by 74 K

The heat capacity of helium = 20.8 J/mol*K

Molecular weight for helium = 4 g/mol

<u>Step 2:</u> Calculate moles of Helium

Moles of Helium = mass of Helium / molar mass of helium

Moles of helium = 250 grams / 4g/mol

Moles of helium = 62.5 moles

<u>Step 3:</u> Calculate heat

Q = n*c*ΔT  or

⇒ with Q = the heat released in Joule

⇒ with n = moles of helium = 62.5 moles

⇒ with c = the heat capacity of helium = 20.8 J/mol*K

⇒ with ΔT = the change in temperature = T2 - T1 = 74 K

Q = 62.5 mol * 20.8 J/mol*K * 74 K

Q = 96200 J = 96 kJ

Since the temperature is raised, this is an exothermic reaction. The heat is released.

There is 96200 J or 96 kJ of heat released.

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Elena-2011 [213]

Answer:

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

Explanation:

CaCO_3(s)\rightarrow CaO(s) + CO_2(g)

Annual production of CaO = 8.6\times 10^{10} kg=8.6\times 10^{13} g

Moles of CaO :

\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with  1.53\times 10^{12} moles of CaO moles of carbon-dioxide moles produced will be:

\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles of carbon-dioxide

Mass of 1.53\times 10^{12} moles of carbon-dioxide:

1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

6 0
3 years ago
The crack shown in the lithosphere was created by the movement of two or more tectonic plates. What is this crack called
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Answer:

The lithosphere is made up of pieces of tectonic plates. These plates are constantly changing and move towards the mantle. Non-stop movement of tectonic plates causes stress on the earth's outermost layer i.e,the crust. When these stresses extends it leads to cause cracks called faults.

Explanation:

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3 years ago
The balanced redox reactions for the sequential reduction of vanadium are given below.
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Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)

2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)

On adding above equations, we get the following combined equation.

2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

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We need 0.0392 grams of Zn metal to completely reduce vanadium.

6 0
3 years ago
Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera
ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

Q=m*Ce*ΔT

Q= heat  (unknown)

m= mass  (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature  (2.75 °C)  

Latent heat:

ΔE=∝mΔHvap

ΔE= latent heat

m= mass  (unknown)

∝= mass fraction (unknown)

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Since Q and E are equal, we can match both equations:

m*Ce*ΔT=∝*m*ΔHvap

Mass fraction is:

∝=\frac{Ce*ΔT}{ΔHvap}

∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}

∝=0,0051

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