Which chemical symbols will complete the equation for this decomposition reaction?

This is the chemical formula for methyl acetate: CH32CO2. Calculate the mass percent of hydrogen in methyl acetate. Round your a

WITCHER [35]

Answer: The mass percent of hydrogen in methyl acetate is 8 %

Explanation:

The given chemical formula of methyl acetate is $CH_3COOCH_3$

To calculate the mass percentage of hydrogen in methyl acetate, we use the equation:

$\text{Mass percent of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of methyl acetate}}\times 100$

Mass of hydrogen = (6 × 1) = 6 g

Mass of methyl acetate = [(3 × 12) + (6 × 1) + (2 × 16)] = 74 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen}=\frac{6g}{74g}\times 100=8.10\%=8\%$

Hence, the mass percent of hydrogen in methyl acetate is 8 %

8 0

2 years ago

In a laboratory setting, concentrations for solutions are measured in molarity, which is the number of moles per liter (mol/L).

slamgirl [31]

Answer:

im pretty sure its A or C im leaning more toward A tho

Explanation:

6 0

1 year ago

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Why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.

drek231 [11]

Net Primary Productivity ... the amount of biomass present in an ecosystem at a particular time .... Explain why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.

8 0

2 years ago

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce

sergeinik [125]

Answer:

The empirical formula is = $C_4H_8O$

The formula of Valproic acid = $C_8H_{16}O_2$

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.00922 = 0.01844 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.403 g /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.009157 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g

Given that the Valproic acid only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C - Mass of H

Mass of the sample = 0.165 g

Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.036412 / 15.999 = 0.002276 moles

Taking the simplest ratio for H, O and C as:

0.01844 : 0.002276 : 0.009157

= 8 : 1 : 4

The empirical formula is = $C_4H_8O$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,

Molecular mass = n × Empirical mass

144 = n × 72

⇒ n = 2

The formula of Valproic acid = $C_8H_{16}O_2$

7 0

2 years ago

Calculate the entropy change for a process in which 3.00 moles of liquid water at 08c is mixed with 1.00 mole of water at 100.8c

kupik [55]

The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:

ΔS = nCln(T₂/T₁)

n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature

We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:

[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K

Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:

ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K

ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K

Now we combine the entropy change of each portion of water to get the total entropy change for the system:

ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K

The entropy change for combining the two temperatures of water is 2.9 J/K.

3 0

2 years ago

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