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evablogger [386]
3 years ago
9

Which of the following compounds are held together by ionic bonds? Select all that apply.

Chemistry
1 answer:
Sergeu [11.5K]3 years ago
7 0

Answer:

Explanation:

All of above except carbon dioxide

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ILL MARK BRAINIEST WHOEVER ANSWERS CORRECTLY
natulia [17]

Answer:

I think its A

sorry if im wrong

Explanation:

4 0
2 years ago
1. A gas having the following composition is burnt under a boiler with 50% excess air.
jeka94

The composition of the stack gas are :

CH_4= 0.8713

C_3H_8 = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of CH_4 : 70% of 100 g = 70 g

Mass of C_3H_8 : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of CH_4 : \frac{70 g}{16.04 g/mol} = 4.3 mole

Number of moles of C_3H_8: \frac{15 g}{144.1 g/mol} = 0.10 mole

Mass of CO : \frac{15 g}{28.01 g/mol} = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of CH_4: \frac{4.3}{4.935} = 0.8713

Mole fraction of C_3H_8 : \frac{0.10}{4.935} = 0.0202

Mole fraction of CO : \frac{0.53}{4.935} = 0.107

Hence, composition of the stack gas are:

CH_4 = 0.8713

C_3H_8 = 0.0202

CO = 0.107

Learn more about mole fraction here:

brainly.com/question/13135950

#SPJ1

8 0
2 years ago
What is the nuclear equation for uranium 235
anzhelika [568]
(1,0)n +(235,92)U --->(91,36)Kr + (142,56) Ba + 3(1,0)n
8 0
3 years ago
If a saturated solution of potassium chloride is cooled from 80℃ to 50℃,how much precipate is formed
harina [27]
Roughly 39 grams, give or take 1 gram
7 0
2 years ago
Isooctane, C8H18, is the component of gasoline from which the term octane rating derives.
lina2011 [118]

Answer:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) Mass of CO₂ produced annually from this combustion of isooctane gasoline = 1.12 × 10⁵ Kg

c) CO₂ produced from the combustion of the gasoline in a year will occupy 5.632 × 10⁷ L

d) There needs to be a minimum of 1.752 × 10⁷ moles of air and 3.92 × 10⁸ L of air for the oxygen to be in excess all through the year of gasoline combustion.

Explanation:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) C₈H₁₈ has a density of 0.792 mg/L.

Since density = mass/volume;

mass = density × volume

Mass of C₈H₁₈ with 4.6 x 10^10 L volume = 0.792 × 4.6 x 10^10 = 3.643 × 10^10 mg = 3.643 × 10⁷ g.

To obtain the mass of CO₂ produced, we need the number of moles of C₈H₁₈ that burned.

Number of moles = mass/molar mass

Molar mass of C₈H₁₈ = (8×12) + 18 = 114g/mol

Number of moles of C₈H₁₈ = (3.643 × 10⁷)/114 = (3.2 × 10⁵) moles.

From the chemical reaction,

1 mole of C₈H₁₈ burns to give 8 moles of CO₂

(3.2 × 10⁵) moles will give 8 × 3.2 × 10⁵ = (2.56 × 10⁶) moles of CO₂

Mass of CO₂ produced = number of moles × Molar mass

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ produced = 2.56 × 10⁶ × 44 = 1.12 × 10⁸ g = 1.12 × 10⁵ kg

c) 1 mole of any gas at stp occupies 22.4L

2.56 × 10⁶ moles of CO₂ will occupy 2.56 × 10⁶ × 22.4 = 5.632 × 10⁷ L

d) 1 mole of C₈H₁₈ requires 23/2 moles of O₂ for complete combustion yearly.

3.2 × 10⁵ moles would require 3.2 × 10⁵ × 23/2 = 3.68 × 10⁶ moles of O₂

O₂ makes up 21% of the air

That is,

0.21 moles of O₂ would be contained in 1 mole of air

3.68 × 10⁶ moles of O₂ would be contained in (3.68 × 10⁶ × 1)/0.21 = 1.752 × 10⁷ moles of air.

1 mole of any gas at stp occupies 22.4L

1.752 × 10⁷ of air will occupy

1.752 × 10⁷ × 22.4/1 = 3.92 × 10⁸ L of air!

3 0
3 years ago
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