I can be found in the 4th period on the periodic table.

If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef

andrezito [222]

Answer:

B

Explanation:

Recall the law of effusion:

$\displaystyle \frac{r_1}{r_2} = \sqrt{ \frac{\mathcal{M}_2}{\mathcal{M}_1} }$

Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.

Let the rate of oxygen be r₁ and the rate of hydrogen be r₂.

The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.

Substitute and solve for r₂:

$\displaystyle \begin{aligned} \frac{(0.5\text{ mol/s})}{r_2} & = \sqrt{\frac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}} \\ \\ r_2 & = \frac{0.5\text{ mol/s}}{\sqrt{\dfrac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}}} \\ \\ & = 2.0\text{ mol/s}\end{aligned}$

Because there are 5 moles of hydrogen gas:

$\displaystyle 5.0\text{ mol} \cdot \frac{1\text{ s}}{2.0\text{ mol}} = 2.5\text{ s}$

In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.

Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.

6 0

3 years ago

Which of the following samples will have the lowest pressure if they are all at the same temperature and in identical containers

snow_lady [41]

Answer:

c. 15 g Kr

Explanation:

The amount of a gas (Moles) is directely proportional to its pressure. That means the higher amount of moles, the highest pressure and vice versa.

Using molar mass of the compounds (Ne=20.2g/mol, Ar = 39.9g/mol, Kr = 83.8g/mol, CO₂ = 44 g/mol and F₂ = 38.0g/mol), moles of 15.0g of each gas are:

Ne = 15g ₓ (1mol / 20.2g) = 0.74 moles of Ne

Ar = 15g ₓ (1mol / 39.9g) = 0.38 moles of Ar

Kr = 15g ₓ (1mol / 83.8g) = 0.18 moles of Kr

CO₂ = 15g ₓ (1mol / 44g) = 0.34 moles of CO₂

F₂ = 15g ₓ (1mol / 38g) = 0.39 moles of F₂

<h3>As 15g of Kr contains the less quantity of moles, this sample will con have the lowest pressure</h3>

8 0

3 years ago

What is the noble gas configuration for telleurium

Maru [420]

[Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.

<h3>What is telleurium?</h3>

Tellurium is a noble gas element that is non-reactive in nature due to complete outermost shell. It has atomic number of 52 which means that it has 52 number of electrons.

So we can conclude that [Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.

Learn more about noble gas here: brainly.com/question/13715159

#SPJ1

4 0

2 years ago

Solid carbon (C) can burn in oxygen (O2). Select

babunello [35]

Answer:

OCO

Another way of writing CO₂

Explanation:

A reaction equation has reactants on the left and products on the right.

The reactants are carbon and oxygen. The product is carbon dioxide.

C + O₂ → CO₂

You might see the equation both ways.

C + O₂ → OCO

C + O₂ in the products would mean no reaction has occurred. The problem can solid carbon can burn in oxygen, so a reaction will occur. For no reaction, you would put "NR" in the products.

OCO is the structural way of writing CO₂. Both have one carbon atom (C) and two oxygen atoms (O).

C + 2O is not possible. Oxygen, if alone, has to be at least O₂ because it's a diatomic molecule.

3 0

3 years ago

What is the chemical name of this compound? Na2S:

ahrayia [7]

Answer:

Sodium Sulfuret

7 0

3 years ago