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Nataly_w [17]
3 years ago
13

What is the coefficient for H2O when the equation ? Ca(OH)2(aq) + ? H3PO4(aq) → ? Ca3(PO4)2(s) + ? H2O(ℓ) is balanced using the

smallest possible integers?
Chemistry
1 answer:
kari74 [83]3 years ago
7 0

Answer:

Coefficient of H_{2}O in the balanced equation with smallest possible integer is 6.

Explanation:

Unbalanced equation: Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)

Balance Ca: 3Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)

Balance PO_{4}  : 3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)

Balance H and O: 3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)

Balanced equation: 3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)

So coefficient of H_{2}O in the balanced equation with smallest possible integer is 6.

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Answer:

Boiling is 212* Freezing is 32*

Explanation:

It simple facts

5 0
3 years ago
100.0 g of 4.0°C water is heated until its temperature is 37.0°C. If the specific heat of water is 4.184 J/g°C, calculate the am
Lady bird [3.3K]

Answer:

13.8072 kj

Explanation:

Given data:

Mass of water = 100.0 g

Initial temperature = 4.0 °C

Final temperature = 37.0°C

Specific heat capacity = 4.184 j/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 37.0°C -  4.0 °C

ΔT = 33.0°C

Q = 100.0 g ×4.184 j/g.°C × 33.0°C

Q = 13807.2 j

Joule to KJ:

13807.2 j  × 1kj  /1000 j

13.8072 kj

5 0
3 years ago
When the equation below is balanced, what is the coefficient for oxygen?<br> C3H8+O2=CO2+H2O
Nataly [62]
Se for no balanceamento e 5. 
5 0
3 years ago
The atomic number is based on the number of protons.
chubhunter [2.5K]
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5 0
3 years ago
How many kilojoules are absorbed when 0.772 mol of N2(g) reacts? N2(g) + 2NO(g) → 2N2O(g) ΔH = 73.8 kJ
ANEK [815]

Answer:

56.97kJ

Explanation:

1 mole of N2 reacts with 73.8kJ

0.772 mole of N2 reacts with xkJ

cross multiply

x= 0.772×73.8

=56.97kJ

5 0
3 years ago
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