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Korolek [52]
3 years ago
11

_Al(NO3)3 +_(NH4)3 P04_AlPO4+_ NH, NO3balancing equations​

Chemistry
1 answer:
astra-53 [7]3 years ago
4 0

Answer:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

The coefficients are: 1, 1, 1, 3

Explanation:

__Al(NO₃)₃ + __(NH₄)₃PO₄ —> __AlPO₄ + __NH₄NO₃

The above equation can be balance as illustrated below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + NH₄NO₃

There are 12 atoms of H on the left side and 4 atoms on the right side. It can be balance by writing 3 before NH₄NO₃ as shown below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

Now the equation is balanced.

The coefficients are: 1, 1, 1, 3

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xeze [42]
Spontaneous reaction.


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3 years ago
In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba ( OH ) 2 was added to 60.0 mL of 0.600 M HCl . The reaction caused the
densk [106]

Answer:

Q sln = 75.165 J

Explanation:

a constant pressure calorimeter:

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∴ m sln = m Ba(OH)2 + m HCl

∴ molar mass Ba(OH)2 = 171.34 g/mol

∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol

⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g

∴ molar mass HCl = 36.46 g/mol

∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol

⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g

⇒ m sln = 3.084 g + 1.313 g = 4.3966 g

specific heat (C):

∴ C sln = C H2O = 4.18 J/g°C

∴ ΔT = 26.83°C - 22.74°C = 4.09°C

heat absorbed (Q):

⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)

⇒ Q sln = 75.165 J

8 0
3 years ago
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Usimov [2.4K]

Answer:

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Explanation:

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2 years ago
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Grease and non-polar too, so the <span>tetrachloroethylene would be a solvent.</span>
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