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Pani-rosa [81]
3 years ago
6

Blance equation __CaBr2 (aq) + ___Li3PO4(aq) → ___Ca3(PO4)2(s) + ____LiBr (aq)

Chemistry
1 answer:
QveST [7]3 years ago
4 0

Answer:

3CaBr2 + 2LI3PO4 - > Ca3(PO4) 2 + 6LiBr

Explanation:

The first one I did was PO4. There are two on the right side, so I added 2 to Li3PO4 on the other side. That balanced the PO4s and then gave me 6 Lithiums so I balanced that one next on the right side. I added 6 to LiBr which balanced the Li but then gave me 6 Br, so I finished it off by adding 3 in front of CaBr2 which balanced the calcium and bromines.

Here was the process:

CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr

Balances PO4 (2on both sides)

CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Lithiums (6 on each side)

3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Calciums and Bromines (3 Calciums and 6 Bromines each side)

Hope this helped!

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Explanation:

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HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxide serve in this reaction
RSB [31]

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Radical chain initiator

Explanation:

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Uncooked lean ground beef can contain up to 12% fat by mass. How many grams of fat would be contained in 0.97 lbs of ground beef
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8 0
3 years ago
The following information is given for benzene, C6H6, at 1atm: boiling point = 80.1 °C Hvap(80.1 °C) = 30.7 kJ/mol specific heat
user100 [1]

<u>Answer:</u> The heat required for the process is 4.24 kJ

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of benzene = 24.8 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

\text{Moles of benzene}=\frac{24.8g}{78.11g/mol}=0.318mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = ?

n = number of moles = 0.318 moles

\Delta H_{rxn} = enthalpy change of the reaction  = 30.7 kJ/mol

Putting values in above equation, we get:

30.7kJ/mol=\frac{q}{0.318mol}\\\\q=(30.7kJ/mol\times 0.318mol)=4.24kJ

Hence, the heat required for the process is 4.24 kJ

6 0
3 years ago
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