Can you Answer the question below

The molar mass of barium nitrate (ba(no3)2) is 261.35 g/mol. what is the mass of 5.30 × 1022 formula units of ba(no3)2? 0.0900 g

liberstina [14]

Molar mass is the ratio of the mass to that amount of the substance. The mass of the barium nitrate in the formula unit is 23.0 grams.

The mass of a substance is the product of the molar mass of the compound and the number of moles of the compound.

Given,

Molar mass of barium nitrate = 261.35 g/mol

If, $6.022 \times 10^{23}$ have a mass of 261.35 g/mol then, $5.30 \times 10^{22}$ formula units will have a mass of,

$\begin{aligned}& = \dfrac{261.35 \times 5.30 \times 10^{22}}{6.022\times 10^{23}}\\\\&= 23.0\;\rm gm\end{aligned}$

Therefore, option C. 23.0 gm is the mass of barium nitrate.

Learn more about mass here:

brainly.com/question/24958554

6 0

2 years ago

What is the mass of 1.72 moles of sodium nitrate? Use the periodic table and the polyatomic ion resource. A. 85.0 g B. 91.2 g C.

Mrac [35]

The answer is C. 146g because you add all of the masses of the individual elements and then mulyiply by 1.72 to get your answer.

7 0

3 years ago

Read 2 more answers

Someone please answer i’ll give you brainly.

lina2011 [118]

Let's hope she didn't watch it without me or i will never be speaking to her again :))

4 0

2 years ago

What is the identity of an ion having 46 protons, 42 electrons and 60 neutrons?

Anon25 [30]

Answer:

first u lick it then out some sliva on it then rub it and once the volcano erupts u put it back in ur mouth and blow real hard ok!

Explanation:

repeat this process till the dong bleeds!

3 0

3 years ago

If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0

3 years ago