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4 years ago

5C + 6O2 = ? What will be the product of molecules formed from this equation?

Chemistry

2 answers:

QveST [7]4 years ago

8 0

Answer:

5C + 6O₂ → 5CO + 7/2 O₂

Explanation:

According to the question the reaction is between carbon and oxygen molecule . The reactant side was given as 5 moles of carbon reacting with 6 moles of oxygen molecules. The chemical reaction between carbon and excess oxygen will form carbon dioxide but with limited oxygen it will form carbon monoxide.

Now let us write and balance the actual chemical equation as required from the question. In limited oxygen the reaction will be

The reactant said 5 moles of carbon which is 5 atoms of carbon reacting with 12 atoms of oxygen. The number of atoms of element on both sides of the chemical equation need to be balance . The balanced equation when oxygen is limited is written as

5C + 6O₂ → 5CO + 7/2 O₂

The product will be 5 moles of carbon monoxide and 3.5 moles of oxygen molecules(7 atoms) since the oxygen is limited . The carbon monoxide later reacts with more oxygen to form carbon dioxide.

harina [27]4 years ago

6 0

Answer:5 moles ofCarbonmonoxide and 3.5 moles of oxygen gas.This in combine to yield carbon dioxide.

Explanation:

5C + 6O2----------5CO + 7/2O2.

When carbon combine with oxygen, carbon monoxide is formed first and it later recombine with oxygen to yield carbon dioxide.

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How many moles of silver are in 8.46E24 atoms of silver?

KonstantinChe [14]

Answer:

14.048 moles I believe.

Explanation:

(8.46 * 10^24) / ( 6.022 * 10^23) = 14.048

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3 years ago

Use the balanced equation below to answer the question that follows

natta225 [31]

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3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

At equilibrium, ________. At equilibrium, ________. all chemical reactions have ceased the rate constants of the forward and rev

balandron [24]

Answer:

<em>At equilibrium, the rate of the forward, and the reverse reactions are equal.</em>

Explanation:

In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.

NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.

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3 years ago

HELP!!! The periodic change between night and day is caused by ____________.

charle [14.2K]

Answer:

C. Earth's revolution around the sun

Explanation:

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3 years ago