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jekas [21]
3 years ago
7

Write a word equation for the combustion of propane??? HELP????

Chemistry
1 answer:
frutty [35]3 years ago
8 0

Answer:

1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.

Explanation:

The word equation for the combustion of propane can be obtained from the chemical equation;

      C₃H₈   +    5O₂    →   3CO₂   +   4H₂O  

The word equation is therefore:

 1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.

For such a combustion reaction, carbon dioxide and water are produced in the process.

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An average microwave uses a frequency of 2.54 GHz. How much energy is emitted?
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Answer:

0.120 m

Explanation:

What you need to know here is that frequency and wavelength have an inverse relationship as described by the equation

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Give reason brick is a solid
Elenna [48]
The brick is solid because it's not hollow or containing spaces or gaps
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3 years ago
37.9 grams of an unknown substance undergoes a temperature increase of
jenyasd209 [6]

Answer:

1.023 J / g °C

Explanation:

m = 37.9 grams

ΔT = 25.0*C

H = 969 J

c = ?

The equation relating these equation is;

H = mcΔT

making c subject of formulae;

c = H / mΔT

c = 969 J / (37.9 g * 25.0*C)

Upon solving;

c = 1.023 J / g °C

4 0
3 years ago
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
3 years ago
PLEASE HELP
nata0808 [166]

Answer:

Is this math? Cause as a fourth grader, I can do Algebra, but not this.

Explanation:

7 0
2 years ago
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