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liq [111]
3 years ago
14

The following thermochemical equation is for the reaction of methane(g) with water(g) to form hydrogen(g) and carbon monoxide(g)

. CH4(g) + H2O(g)3H2(g) + CO(g) H = 206 kJ When 4.20 grams of methane(g) react with excess water(g),______ kJ of energy are___________.
Chemistry
1 answer:
Andrei [34K]3 years ago
6 0

Answer:

54.075KJ

Explanation:

Since methane is the limiting reactant.

From the reaction equation,

16 g of methane absorbs 206 KJ of energy

4.20 g of methane will absorb 4.20 × 206/16 = 54.075KJ of energy was absorbed.

From the equation, the value of the enthalpy of reaction is positive, this means that energy is absorbed in the reaction of methane and water. Hence the answer.

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What is the total number of grams of mg consumed when 0.50 mole of h2 is produced
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4 0
2 years ago
1. The following reaction does not proceed to form a product: H2O + Au---&gt; no reaction. Why is that?
Xelga [282]
1. B
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2. In order to balance the equation, you must sure there are equal moles of each element on the left and right side of the equation:
2C₂H₆ + 7O₂ → 4CO₂ + ₆H₂O

3. The number of moles of sodium atoms on the left of the equation must be equal to the number of moles of sodium atoms on the right, as per the law of conservation of mass. The answer is B.

4. C.
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8 0
3 years ago
Read 2 more answers
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
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