From the Graham's law of effusion;
R1/R2 = √MM2/√MM1
Molar mass of chlorine gas is 71
Therefore;
1.87= √ 71 /√mm1
= 1.87² = 71/mm1
mm1 = 71/1.87²
= 71/3.4969
= 20.3
Thus, the molar mass of the other gas is 20.3 , and i think the gas is neon

6 0

3 years ago

A student wants to heat a piece of iron so that its temperature rises by 20 degrees C. What information does she need about the

LekaFEV [45]

Mass................

5 0

2 years ago

Read 2 more answers

in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125

PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0

2 years ago

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0

2 years ago