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irina [24]
3 years ago
5

Once at the top. the wax molecules begin to___________ energy to the glass then the surrounding air.

Chemistry
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

release

Explanation:

A that is the answer

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Consider 2 Al + 6 HCl → 2 AlCl3 + 3 H2 , the reaction of Al with HCl to produce hydrogen gas. What is the pressure of H2 if the
Margaret [11]

Answer : The pressure of hydrogen gas is 8.96 atm.

Explanation :

The given balanced chemical reaction is:

2Al+6HCl\rightarrow 2AlCl_3+3H_2

From the balanced chemical reaction we conclude that,

As, 2 moles of Al react to give 3 moles of H_2 gas

So, 4.50 moles of Al react to give \frac{3}{2}\times 4.50=6.75 moles of H_2 gas

Now we have to calculate the moles of H_2 gas when percent yield is 75.4.

\text{The moles of }H_2=75.4\% \times 6.75=\frac{75.4}{100}\times 6.75=5.09moles

Now we have to calculate the pressure of H_2 gas.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of hydrogen gas = ?

V = Volume of the hydrogen gas = 14.0 L

n = number of moles of hydrogen gas = 5.09 moles

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of hydrogen gas = 300 K

Putting values in above equation, we get:

P\times 14.0L=5.09mol\times 0.0821L.atm/mol.K\times 300K\\\\P=8.96atm

Therefore, the pressure of hydrogen gas is 8.96 atm.

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over the last 100 years the number of lives lost during major storms like hurricanes has steadily dropped what could be conclude
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For every 1.0 mole of glycine in the sample, how many molecules of methionine are present? (for help performing calculations wit
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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
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