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2 years ago

School Math. Homework.

Chemistry

1 answer:

Ksju [112]2 years ago

3 0

Answer:

The answer is TRUE!!!

Explanation:

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You are given a sample of an element with ONE valence electron.

Gnoma [55]

Answer:

the amount of valence electron, cna be found in the periodic table, which is called the atomic number which is top of the symbol

hope this helps!!! best fo luck to you

5 0

2 years ago

2.600 * 10 ^ - 5 * 6.100 * 10 ^ - 5 Express in scientific notation

Anna [14]

Answer:

hope its help you

Explanation:

2.600 × 10^-5 × 6.100 × 10^-5 = 1.586 × 10-9

3 0

3 years ago

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

ammonia gas reacts with oxygen gas, o2(g), to produce nitrogen dioxide and water. when 28.5 g of ammonia gas reacts with 83.4 g

jolli1 [7]

The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.

A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)

NH₃ + O₂ → NO₂ + H₂O

The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O

Calculate the number of moles from the reactant

Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol
Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol
n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.

According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.

Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol
The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol

The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.

P = (1.35 ÷ 1.49) × 100%

P = 0.9052 × 100%

P = 90.52%

Learn more about stoichiometry here: brainly.com/question/13691565

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7 0

1 year ago

The metal element cesium has one electron in its outer orbit. Will cesium atoms form positively charged or negatively charged io

r-ruslan [8.4K]

Answer:

Caesium atoms will form positively charged ions.

Explanation:

Due to them having one electron in their outer orbit, it is very likely that they will give that electron away to form an octet and become stable. Hence, since they lose an electron, they lose a negative charge, and in comparison to their former non-ion self, they have gained a positive charge.

Hence, they will form positively charged ions.

Hope this helped!

8 0

2 years ago