This circuit shows a battery and wires connected to a lightbulb. The chemical energy in the battery is converted to:

5.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 7

slega [8]

Answer:

The molecular formula = C6H6

Explanation:

Step 1: Data given

Mass of compound X = 5.00 grams

Mass of products =

CO2 = 16.39 grams

H2O = 3.46 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16. 0g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles CO2 = 16.93 grams / 44.01 g/mol

Moles CO2 = 0.385 moles

Moles C = 1* 0.385 = 0.385 moles

Moles H2O = 3.46 grams / 18.02

Moles H2O = 0.192 moles

Moles H = 2* 0.192 = 0.384 moles

Step 3: Calculate mass

Mass = moles * molar mass

Mass C = 0.385 moles *12.0 g/mol

Mass C = 4.62 grams

Mass H = 0.39 grams

Mass O = 5.00 - 4.62 -0.38 moles

Mass O = 0 grams

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.385 moles / 0.384 = 1

H: 0.384 moles / 0.384 = 1

The empirical formula is CH

This molecular formula is 13 g/mol

We have to multiply the empirical formula by n

n = 78 g/mol / 13 g/mol

n = 6

The molecular formula = 6*(CH) = C6H6

4 0

3 years ago

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

3 years ago

Is carbon tetrafluoride a polar molecule?

atroni [7]

It is non-polar molecule.
CF₄ - dipole moment = 0D

5 0

3 years ago

What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

5 0

3 years ago

The name for the following formula C9H18 should be

Elza [17]

Answer:

cyclononane C9H18 nonane

7 0

3 years ago