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3 years ago

A compound has an empirical formula of CHBr2. If it’s molar mass is 345.6 grams, what is it’s molecular formula

Chemistry

1 answer:

djyliett [7]3 years ago

3 0

Answer:

The molecular formula is C2H2Br4

Explanation:

Molar Mass of CHBr2 = 12 + 1 + (2x80) = 12 + 1 + 160 = 173

The molecular formula = n x empirical formula

Molar Mass of the compound = 345.6

Empirical formula = CHBr2

n(CHBr2) = 345.6

n x 173 = 345.6

n = 345.6/173 = 2

Therefore the molecular formula is n(CHBr2) = 2(CHBr2) = C2H2Br4

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If potassium (K) has a density of 0.86 g/cm3, which element would have a similar density?

Nata [24]

Answer:

C: Sodium

Explanation:

Iron had a way higher density neon has way to know as well as carbon and the closest one is sodium

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3 years ago

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When the valves are opened and the gases are allowed to mix at a constant temperature, what is the distribution of atoms in each

jek_recluse [69]

Answer:

Equal number of atoms of each gas in each container

Explanation:

When the valves opened, the two contaienrs become one and the gases beging to mix by diffusion. This phenomenom is produced by the differeces of concentration of a gas between two points of the container.

The gases will continue diffunding util their concentration in both containers are equal.

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3 years ago

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol$

The given chemical equation follows:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = $\frac{3}{2}\times 0.467=0.7005mol$ of copper

Now, calculating the mass of copper from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

$0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g$

Hence, the theoretical yield of copper is 44.48 grams

4 0

4 years ago

(03.03 MC)

Jet001 [13]

Answer:

- Carbon shares four of its electrons, and each oxygen shares two of its electrons.

Explanation:

Carbon needs 4 electrons to reach a full outer shell while oxygen needs 2 to reach a full outer shell.

7 0

2 years ago

A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

6 0

3 years ago