How many Oxygen atoms in: 5H2SO3

Liquid octane (CH) has a density of 0.7025 g/mL at 20 °C. Find the true mass (murue) of octane when the mass weighed in 18 air i

Goshia [24]

Explanation:

According to Buoyance equation,

m = $[m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]$

where, m = true mass

m' = mass read from the balance = 17.320 g

$d_{a}$ = density of air = 0.0012 g/ml

$d_{w}$ = density of the balance = 7.5 g/ml

d = density of liquid octane = 0.7025 g/ml

Now, putting all the given values into the above formula and calculate the true mass as follows.

m = $[m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]$

= $[17.320 g \times \frac{1 - \frac{0.0012 g/ml}{7.5 g/ml}}{1 - \frac{0.0012 g/ml}{0.7025}}]$

= $17.320 g \times 0.999850$

= 17.317 g

Thus, we can conclude that the true mass of octane is 17.317 g.

7 0

3 years ago

Hydrogen chloride gas and oxygen gas react to form water and chlorine gas. A reaction mixture initially contains 53.2 g of hydro

Sergio [31]

Answer: The mass of the excess reactant (oxygen gas) is 3.136 grams

Explanation:

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

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

For HCl:

Given mass of HCl = 53.2 g

Molar mass of HCl = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of HCl}=\frac{53.2g}{36.5g/mol}=1.46mol$

For oxygen gas:

Given mass of oxygen gas = 26.5 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{26.5g}{32g/mol}=0.828mol$

The chemical equation for the reaction of HCl and oxygen gas follows:

$2HCl+O_2\rightarrow H_2O+Cl_2$

By Stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of oxygen gas

So, 1.46 moles of HCl will react with = $\frac{1}{2}\times 1.46=0.73mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, HCl is considered as a limiting reagent because it limits the formation of product.

Excess moles of oxygen gas = (0.828 - 0.73) = 0.098 moles

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

Molar mass of oxygen gas = 32 g/mol

Excess moles of oxygen gas = 0.098 moles

Putting values in equation 1, we get:

$0.098mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.098mol\times 32g/mol)=3.136g$

Hence, the mass of the excess reactant (oxygen gas) is 3.136 grams

4 0

3 years ago

A sample of Br 2 ( g ) Br2(g) takes 42.0 min 42.0 min to effuse through a membrane. How long would it take the same number of mo

Simora [160]

15 mins

~Savannah

Jokes from Savannah~

A man goes to the doctor, concerned about his wife’s hearing. The doctor says, “Stand behind her and say something and tell me how close you are when she hears you.”

The man goes home, sees his wife in the kitchen, cutting carrots on the countertop. About 15 feet away he says, “Honey, what’s for dinner?” Nothing. He gets halfway to her and repeats the same question. Nothing. Very concerned, he gets right behind her and asks again “What’s for dinner?”

She turns around and says “For the THIRD time, beef stew!”

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Q: What do you get when you cross a vampire with a snowman?

A: Frostbite!

6 0

3 years ago

How many atoms are found in a 15.5 g sample of bismuth?

dmitriy555 [2]

Answer:

The number of particles in the 41.8g sample of Bismuth is 12.044 × 10²³

Explanation:

3 0

3 years ago

When 280. mL of 1.50 × 10⁻⁴ M hydrochloric acid is added to 135 mL of 1.75 × 10⁻⁴ M Mg(OH)₂, the resulting solution will be?

goldfiish [28.3K]

Answer:

B is the correct answer!

Explanation:

Balanced equation is Mg(OH)2 + 2 HCl = MgCl2 + 2 H2O

HCl is the limiting reactant because it gives the smaller amount of MgCl2 I did the math. This means you will be left with Mg(OH)2 since HCl completely runs out.

If HCl is the limiting reactant, Mg(OH) 2 is the excess reactant.

Mg(OH) 2 is a base, so the solution will be basic.

6 0

3 years ago