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Slav-nsk [51]
2 years ago
11

How many moles of H3PO4 would produce 12 moles of water in this reaction? ЗKOH + H3PO4 K3PO4 + 3H20

Chemistry
2 answers:
Blizzard [7]2 years ago
7 0

Answer:4 moles of H₃PO₄

Explanation: did the test homie

Olenka [21]2 years ago
4 0

Answer:

4 moles of H₃PO₄

Explanation:

The reaction expression is given as;

     3KOH   + H₃PO₄   →   K₃PO₄   +  3H₂O

Number of moles of water  =  12moles

Unknown:

Number of moles of H₃PO₄ = ?

Solution:

 From the balanced reaction expression we see that;

          3 moles of water is produced from 1 mole of H₃PO₄

So;    12 moles of water would be produced from \frac{12}{3} = 4 moles of H₃PO₄

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How much heat is required to change the temperature of a 15 g aluminum can with 100 g of water from 24.5°C to 55°C?
Assoli18 [71]
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium 
Heat gained by water = 100 × 4.186 × 30.5 
                                   = 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5 
                                          = 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
3 0
3 years ago
When a 500. gram sample of water at 19.0°C absorbs 8400 Joules of heat, the temperature of the water
Westkost [7]

Answer: 23.0ºC

Explanation:

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7 0
2 years ago
What is the molality of a solution prepared by dissolving 150.0 g C6H12O6 in 600.0 g of H2O?
elena-14-01-66 [18.8K]

Answer:

50/36 = 25/18

Explanation:

Solution at attachment box

Molality = mole of dissolvable (this question glucose) / kg of water

4 0
3 years ago
Read 2 more answers
How many grams of water are needed to dissolve 27.8 g of ammonium nitrate?
andreyandreev [35.5K]
The moles of ammonium nitrate needed to dissolve 0.35 moles
The moles of water that will react is 0.35 moles as due to ratio
so mass of water will be 0.35 x 18=6.3g 
                                 MASS OF WATER WILL BE 6.3 g
7 0
3 years ago
In the laboratory a student determines the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it i
fgiga [73]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc = 98.37 °C

Initial temperature of water = 24.16 °C

Final temperature of water (and zinc) = 25.70 °C

Specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate Specific heat of zinc

Heat lost by zinc = heat won by water

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

Qzinc = -Qwater

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

7 0
3 years ago
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