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3 years ago

Im hella confused and need some help

Chemistry

1 answer:

tatiyna3 years ago

6 0

An oxide is a molecule that has 2 Oxygen atoms in its empirical formula. Stoichiometry would be used as the ratio of Oxygen to Dioxides is 3:2. So 52.29 would be multiplied by 3/2. The answer is 78.435 Mol of CO2 are formed.

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What is the answer to 4x - 16

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It simply equals 4x-16.......
the expression is in lowest terms, so you can’t do anything else

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All of the factors described below could account for the difference in the sizes of the tomatoes. Which describes a way in which

disa [49]

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D my guy

Explanation:

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3 years ago

A chemist measures the amount of chlorine gas produced during an experiment. She finds that of chlorine gas is produced. Calcula

Trava [24]

Answer:

0.779 moles.

Explanation:

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The diagram below represents a sodium atom bonding to a chlorine atom to form sodium chlorine.

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3 years ago

3mL of cyclohexanol (density = 0.9624 g/mL, Molecular weight = 100.158 g/mol) reacts with excess sulfuric acid to produce cycloh

TiliK225 [7]

Answer:

$n_{C_6H_{10}}=0.03molC_6H_{10}$

Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

$n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}$

Besides, the mass could be computed as well by using the molar mass of cyclohexene:

$m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}$

Even thought, the volume could be also computed by using its density:

$V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}$

Best regards.

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3 years ago

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