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madreJ [45]
3 years ago
12

Which of the following is the correct skeleton equation for the reaction that takes place when solid phosphorus combines with ox

ygen gas to form diphosphorus pentoxide?
A. P(s) + O2(g) --> P2O5(s)

B. P2O5(s) --> P2(s) + O2(g)

C. P(s) + O(g) --> P5O2(g)

D. P(s) + O2(g) --> PO2(g)
Chemistry
1 answer:
Aliun [14]3 years ago
4 0

Answer: Option (A) is correct.

Explanation:

In diphosphorus pentoxide there will be two phosphorous atoms and 5 oxygen atoms according to its name.

Therefore, the formula for diphosphorus pentoxide will be P_{2}O_{5} and it is a solid.

Thus, the correct skeleton equation for the reaction that takes place when solid phosphorus combines with oxygen gas to form diphosphorus pentoxide will be as follows.

                 P(s)+ O_{2}(g) \rightarrow P_{2}O_{5}(s)


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ryzh [129]
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When water changes state from a liquid to a solid is energy absorbed or lost?
Brrunno [24]

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Loses

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Calculate the number of grams of Hydrogen required to produce 5.73 grams of water. 2H2+O2--->2H2O 2 grams 0.636 grams 1.2 gra
rusak2 [61]

Explanation:

32

2H

2

+O

2

→2H

2

O

Molecular mass of H

2

=2 g/mol

Molecular mass of O

2

=32 g/mol

From the balanced chemical equation,

2×2=4 g of hydrogen requires 32 g of Oxygen to react completely

3 0
3 years ago
A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A
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Answer:

T_{2}=16,97^{\circ}C

Explanation:

The specific heats of water and steel are  

Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}

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Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium T_{2}_{w}= T_{2}_{s}  

An energy balance can be written as

m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}  

Replacing

1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)

Then, the temperature T_{2}=16,97^{\circ}C

8 0
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