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jok3333 [9.3K]
2 years ago
9

Please help I will mark brainy

Chemistry
1 answer:
Artyom0805 [142]2 years ago
3 0

Answer:

the answer is c I took the test

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What state of matter change is rain?
Charra [1.4K]

Answer:

The state of matter of rain is liquid. Clouds are 2 states of matter, liquid and gas

Explanation:

5 0
2 years ago
How many moles of O2 is needed to form 2.19 moles of CO2 in the reaction
bagirrra123 [75]
It is always half of what the o2 is
6 0
3 years ago
2<br> CuCl2 +4KI -&gt; 2 Cul + 4 KCI + 12
Amanda [17]

Answer:

Explanation:

This is an example of a limiting reactant question, and is very common as a general chemistry problem.

We first see the balanced equation, that is:

2CuCl2+4KI→2CuI+4KCl+I2

We first need to find the limiting reactant

We see that 0.56 g of copper(II) chloride (CuCl2) reacts with 0.64 g of potassium iodide (KI) . So, let's convert those amounts into moles.

Copper(II) chloride has a molar mass of

134.45 g/mol . So in 0.56 g of copper(II) chloride, then there exist

0.56g134.45g/mol≈4.17⋅10−3 mol

Potassium iodide has a molar mass of

166 g/mol . So, in 0.64 g of potassium iodide, there exist

if it wrong i am sorry

5 0
3 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
What is the gram formula mass of Na2CO3 multiplied by 10H2O
Lerok [7]
Enter a chemical formula to calculate its molar mass and elemental composition: Notice: your ... as a zero '0' Molar mass of Na2Co3*10H2O<span> is 402.9319 g/mol ...</span>
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