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7.02 x 10^23 molecules of X2, a ssubstance consisting of diatomic molecules, has a mass of 296 grams. Determine the atomic weigh

Fiesta28 [93]

Answer:

d. 127 g/mol.

Explanation:

Hello!

In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:

$mol=7.02x10^{23}molec*\frac{1mol}{6.022x10^{23}molec}=1.17mol$

Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:

$MM=\frac{296g}{1.17mol}\\\\MM=254g/mol$

It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.

Best regards!

4 0

2 years ago

Which of the following is the best explanation for the fact that crushing increases the rate at which a solid dissolves in a liq

Oksana_A [137]

It’s D cause i know I know I know I know

7 0

2 years ago

Find the normality of 0.321 g sodium carbonate in a 250 mL solution.

Radda [10]

Answer:

the normality of the given solution is 0.0755 N

Explanation:

The computation of the normality of the given solution is shown below:

Here we have to realize the two sodiums ions per carbonate ion i.e.

N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)

= 0.1886eq ÷ 0.2500L

= 0.0755 N

Hence, the normality of the given solution is 0.0755 N

5 0

2 years ago

What is the chemical name for the compound with the formula Na2S?

Marina86 [1]

I think it's Sodium Sulfide, because Na2S contains one Sodium and two Sulfur. Plus Sulfide<span> is used to describe any of three types of chemical compounds that contain sulfur.</span>

3 0

3 years ago

Read 2 more answers

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

2 years ago