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Flauer [41]
2 years ago
4

Write a balanced equation for the combustion of 1 mol of C4H10O(l). Express your answer as a chemical equation. Identify all of

the phases in your answer.
Chemistry
2 answers:
leonid [27]2 years ago
6 0

Answer:

C4H10O(l) + 6O2(g) → 4CO2(g) + 5H2O(g)

Explanation:

Step 1: Data given

Combustion of C4H10O means adding oxygen gas

Step 2: Balancing the equation

C4H10O(l) + O2(g) → CO2(g) + H2O(g)

On the left side we hace 4x C, on the right side we have 1x C. To balance the amount of C, we have to multiply CO2 (on the right) by 4.

C4H10O(l) + O2(g) → 4CO2 (g)+ H2O(g)

On the left side we have 10x H, on the right side we have 2x H. To balance the amount of H, we have to multiply H2O (on the right side) by 5.

C4H10O(l) + O2(g) → 4CO2(g) + 5H2O(g)

On the left side we 3x O (1x in C4H10O and 2x in O2) on the right side we have 13x O (8x in CO2 and 5x in H2O). To balance the amount of o on both sides, we have to multiply O2 (on the left side) by 6.

Now the equation is balanced.

C4H10O(l) + 6O2(g) → 4CO2(g) + 5H2O(g)

marshall27 [118]2 years ago
4 0

Answer:

C4H10O + 6O2 —> 4CO2 + 5H20

Explanation:

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A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution
mihalych1998 [28]

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= \frac{4.00 g}{245 g/mol}=0.01633 mol

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

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Let the molarity of the solution be M_1

M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M

A students then takes 4 .00 mL of M_1 solution and dilute it to 275 ml.

M_1=0.1633 M

V_1=4.00 mL

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V_2=275 mL (after dilution)

M_1V1=M-2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M

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Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M

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Molarity of the nitrate ions = [NO_3^{-}]=0.004750 M

[NO_3^{-}]=\frac{n}{0.275 L}

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

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Hydrogen sulfide (H2S) is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is to treat the wa
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Explanation:

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