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morpeh [17]
2 years ago
7

How much heat in calories must be added to change 100.0 grams of water from 19.7 - 87 9°C?

Chemistry
1 answer:
anyanavicka [17]2 years ago
3 0

Answer:

6820 calories

Explanation:

Specific heat of water =  1 c / gm-C

100 gm * (87.9-19.7) C *  1 c/gm-C = 6820 cal

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b

Explanation:

because motions make things hotter

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Where are blood cells produced? A. Red Cross B. Bones C. Digestive system D. Cirulatory system
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B.Bones
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The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr
jenyasd209 [6]

Answer : The final temperature of the water is, 22.5166^oC

Explanation : Given,

Mass of benzene = 8.600 g

Molar mass of benzene = 78 g/mole

First we have to calculate the moles of benzene.

\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives \frac{6542 kJ}{2}\times 0.1103=360.7913kJ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})

where,

q_w = heat released = 360.7913 kJ = 36079.13 J

m_w = mass of water = 5691 g

c_w = specific heat of water= 4.18J/g^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 21^oC

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

36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)

T_{final}=22.5166^oC

Therefore, the final temperature of the water is, 22.5166^oC

8 0
4 years ago
What is the correct equation for water gas​
garik1379 [7]

Answer:

Water-gas A mixture of carbon monoxide (CO) and hydrogen (H2) produced by passing steam over red-hot coke using the endothermic reaction C + H2O # CO + H2.

Explanation:

7 0
3 years ago
Two gas-containers, A and B, are connected with a valve. The first container, A, has a volume of 135 mL, and the second containe
stiv31 [10]

Answer:

85.5 mmHg is the pressure of the gas sample when the valve is opened.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas in container A = 165 mmHg

P_2 = final pressure of gas = ?

V_1 = initial volume of gas in container A= 135 mL

V_2 = final volume of gas = 135 mL + 117 mL = 252 mL

T_1 = initial temperature of gas in container A = 22.5^oC=273+22.5=295.5 K

T_2 = final temperature of gas = 12.7^oC=273+12.7=285.7K

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

P_2=\frac{P_1V_1\times T_2}{T_1\times V_2}

=\frac{165 mmHg\times 135 mL\times 285.7 K}{295.5 K\times 252 mL}

P_2=85.5 mmHg

85.5 mmHg is the pressure of the gas sample when the valve is opened.

8 0
4 years ago
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