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ratelena [41]
3 years ago
7

What mass in grams of PCl5 will be produced from 0.25 moles of PCl3 and excess chlorine gas?

Chemistry
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer:

d

Explanation:

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Given the following equation, what is the correct form of the conversion factor needed to convert the number of moles O2 to the
jek_recluse [69]

Answer:

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Explanation:

Step 1: Data given

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate the mol ratio

For 3 moles O2 we'll have 4 moles Fe

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Option b says we have 2 mole Fe2O3 for each 4 moles Fe

This doesnt say anything about O2. So doesn't apply for this question.

Option C says we have 4 moles of Fe for each 2 moles Fe2O3

This is the same as option B, so doesn't apply for this question.

Option D says for each 3 moles of O2 we have 2 Fe2O3

This is true, but doesn't say anything about Fe so doesn't apply here.

4 0
3 years ago
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Which distinguishes an atom of one element from an atoms of a different element?
dalvyx [7]

Answer:

The number of protons you welcome

Explanation:

7 0
3 years ago
Convert 19 kcal into joules
il63 [147K]

19~ \text{kcal} = 19 \times 1000 ~\text{cal } = 19000~ \text{cal} = 19000 \times 4.184~ \text{J} = 79496~ \text{J}

7 0
2 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
3 years ago
There are three ways to give an object a static charge. They are friction, conduction and induction. Two pieces of cellophane ta
kotykmax [81]

Answer:

b) The two pieces of tape will repel as both have obtained a static charge.

this is the right answer.

Explanation:

5 0
3 years ago
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