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3 years ago

How did they get the limiting reactant as O2???

Chemistry

1 answer:

vredina [299]3 years ago

5 0

Better picture? you ksksmskakakaoa

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How do evaporation and condensation relate to heat

ale4655 [162]

Evaporation and condensation is caused by heat. That is how they are related. Heat causes the liquid to evaporate and float into the atmosphere which causes the formation of clouds.

Hope this helped. Have a great day!

7 0

4 years ago

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A gram of gasoline produces 45.0kJ of energy when burned. Gasoline has a density of 0.77/gmL. How would you calculate the amount

Ber [7]

Answer:

Math expression: $=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ$

Explanation:

Given:

Energy produced per gram of gasoline = 45.0 kJ

Density of gasoline = 0.77 g/ml

Volume of gasoline = 5.2 L=5200 ml

To determine:

The amount of energy produced by burning 5.2 L gasoline

Calculation set-up:

1. Calculate the mass (m) of gasoline given the density (d) and volume (v)

$m = d*v\\\\m = 0.77 g/ml*5200 ml$

2. Calculate the amount of energy produced

$=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ=180180 kJ$

7 0

3 years ago

PLEASE HELP!!!

umka21 [38]

Answer:

the answer is A

Explanation:

physical changes does not change the property of matter.

7 0

2 years ago

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What are the factors that affect the rate of dissolution

Leto [7]

The solvent, solute, and solution.

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3 years ago

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When the oxide of generic metal M is heated at 25C, only a negligible amount of M is produced. MO2(s) <---> M(s)+O2(g) del

Valentin [98]

Answer:

1. $MO_2(s)+C(s)CO_2(g)+M(s)$

2. $K=3.57x10^{18}$

Explanation:

Hello,

1.) By coupling the given reaction with the formation of carbon dioxide, one states the total reaction as:

$MO_2(s)M(s)+O_2(g)\\C(s)+O_2(g)CO_2(g)$

____________________________________

$MO_2(s)+C(s)CO_2(g)+M(s)$

2.) Now, since we know that the Gibbs free energy for the decomposition of the metal is 288.5kJ/mol and the Gibbs free energy for the formation of carbon dioxide has a value of −394.39kJ/mol, the total Gibbs free energy for this process is:

Δ $G^o=288.5kJ/mol-394.39kJ/mol=-105.89kJ/mol$

So the equilibrium constant is:

$K=exp(-\frac{DeltaG^0}{RT} )\\K=exp(-\frac{-105890J/mol}{8.314J/molK*298.15K} )\\K=3.57x10^{18}$

Best regards.

3 0

3 years ago