a gas made up of atoms escapes through a pinhole 2.83 times as fast as o₂ gas. write the chemical formula of…

Explanation:

Step1: Recall Graham's law of effusion

According to Graham's law, $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$, where $r_1$ and $r_2$ are the rates of effusion of two gases and $M_1$ and $M_2$ are their molar - masses. Let the unknown gas be gas 1 and $O_2$ be gas 2. We know that $\frac{r_1}{r_2}=2.83$. The molar - mass of $O_2$ is $M_2 = 32\ g/mol$.

Step2: Rearrange Graham's law to solve for $M_1$

From $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$, we can square both sides to get $(\frac{r_1}{r_2})^2=\frac{M_2}{M_1}$. Then, $M_1=\frac{M_2}{(\frac{r_1}{r_2})^2}$. Substitute $\frac{r_1}{r_2}=2.83$ and $M_2 = 32\ g/mol$ into the formula: $M_1=\frac{32}{(2.83)^2}$. Calculate $(2.83)^2 = 8.0089$. Then $M_1=\frac{32}{8.0089}\approx4\ g/mol$.

Step3: Identify the gas

The gas with a molar - mass of approximately $4\ g/mol$ is helium, and its chemical formula is $He$.

Answer:

$He$