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DanielleElmas [232]
4 years ago
15

When the following equation is balanced with the lowest whole number coefficients possible, what is the coefficient in front of

Chemistry
2 answers:
Anastaziya [24]4 years ago
7 0
Given: C3H8(g) + O2(g) ----> CO2 (g) + H2O (g)

Step : Put a 3 in front of CO2 (g) to balance C

=> C3H8(g) + O2(g) ----> 3CO2 + H2O to balance H

Step 2: Put a 4 in front of H2O

=> C3H8 (g) + O2(g) -----> 3CO2 (g) + 4H2O (g)

Step 3: Given that there are 3*2 + 4 = 10 O to the right side, put a 5 in front of O2 to balance O:

=> C3H8(g) + 5O2(g) -----> 3CO2(g) + 4H2O(g)

You can verify that the equation is balanced.

So, the answer is that the coefficient in front of O2 is 5.
neonofarm [45]4 years ago
6 0

<span><span>Reaction stoichiometryLimiting reagent</span><span>CompoundCoefficientMolar MassMolesWeight</span><span>C3H8144.09562</span><span>O2531.9988</span><span>CO2344.0095</span><span>H2O4<span>18.01528</span></span></span><span>Balanced equation:
C3H8 + 5 O2 = 3 CO2 + 4 H2O</span>
<span>Reaction type: combustion
</span>
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If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the mola
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Volume Ba(OH)2 = 23.4 mL in liters : 

23.4 / 1000 => 0.0234 L

Molarity  Ba(OH)2 = 0.65 M

Volume HNO3 = 42.5 mL in liters:

42.5 / 1000 => 0.0425 L

number of moles Ba(OH)2 :

n = M x V

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Mole ratio :

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1 mole Ba(OH)2 ---------------- 2 moles HNO3
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moles HNO3 = 0.01521 x 2 / 1

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= 0.03042 moles HNO3

Therefore:

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3 years ago
For H 3 PO 4 and H 3 BO 3 , does the subscript " of hydrogen in these two formulas seem to result in additional ions in solution
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Answer:

See Explanation

Explanation:

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The next to consider is the type of electrolyte the both solution are; the both solutions are weak electrolytes and weak electrolytes do not ionize to a large extent.

The implication of this is that, not so much number of ions is added to the solution due to the poor ionization of these weak electrolytes. Hence, in spite of the subscript of 3, the conductivity of the solution does not significantly improve for the reason stated here quite unlike when strong electrolytes are used.

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