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4 years ago

Caz(PO4)2 + H2504 - CaSO4+H3PO4

Chemistry

1 answer:

m_a_m_a [10]4 years ago

7 0

Answer:

Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4

The coefficients are 1, 3, 3, 2

Explanation:

Ca3(PO4)2 + H2SO4 —> CaSO4 + H3PO4

From the above equation,

There are 3 atoms of Ca on the left and 1 atom of Ca on the right. To balance Ca, put 3 in front of CaSO4 as shown below

Ca3(PO4)2 + H2SO4 —> 3CaSO4 + H3PO4

Now, we have 3 atoms of SO4 on the right and 1atom on the left. To balance SO4, put 3 in front of H2SO4 as shown below:

Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + H3PO4

Looking closely, there are 6 atoms of H on the left and 3 on the right. Therefore, it is balanced by by putting 2 in front of H3PO4 as shown below:

Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4

The coefficients are 1, 3, 3, 2

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What is the temperature of 0.257 mol of O2 occupying 6.78 L at 0.856 atm?

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From the ideal gas law, PV= n RT

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3 years ago

Understanding the high-temperature formation and breakdown of the nitrogen oxides is essential for controlling the pollutants ge

kakasveta [241]

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$1.143*10^{3} J/mol$

Explanation:

Using the Arrhenius equation for the give problem:

$k = A*exp^{\frac{-E_{a} }{RT} }$

Taking the natural log (ln) of both sides, we have:

$ln k = ln A - \frac{E_{a} }{RT}$

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$ln k_{1} = ln A - \frac{E_{a} }{RT_{1} }$ (1)

$ln k_{2} = ln A - \frac{E_{a} }{RT_{2} }$ (2)

Subtracting equation (1) from equation (2), we have:

$ln k_{2} - ln k_{1} = \frac{E_{a} }{RT_{1} } - \frac{E_{a} }{RT_{2} }$ (3)

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$k_{2} = 0.0815 L/mol-s$ ; $T_{2} = 947°C = 737+273 = 1220 K$

Therefore, equation (3) becomes:

$ln 0.0815 - ln 0.0796 = E_{a}[\frac{1}{8.314*1010} - \frac{1}{8.314*1220}]$

-2.507 - (2.531) = Ea*[0.00012 - 0.000099]

Ea = 0.024/0.000021 = 1142.86 J/mol

The activation energy of the reaction in scientific notation is $1.143*10^{3} J/mol$

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