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Marat540 [252]
3 years ago
14

An automobile gasoline tank holds 22 kg of gasoline. When the gasoline burns, 86 kg of oxygen is consumed and carbon dioxide and

water are produced. What is the total combined mass of carbon dioxide and water produced?
Chemistry
1 answer:
rosijanka [135]3 years ago
7 0
The law of conservation of mass states that mass is an isolated system and is neither created nor destroyed.

Applying this concept on a chemical reaction means that the total mass of reactants should be equal to the total mass of the products.

In this question, we have gasoline and oxygen as reactants and we have carbon dioxide and water as products.
Based on the law, the sum of the masses of oxygen and gasoline is equal to the sum of masses of carbon dioxide and water.
Therefore,
mass of carbon dioxide and water = 22 + 86 = 108 kg
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For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. The reaction is endothermic. What do yo
jeyben [28]

Answer:

The correct answer is : 'the concatenation of NO will increase'.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs.

2NOBr(g)\rightleftharpoons 2NO(g) + Br_2(g)

As, this is an endothermic reaction, increasing temperature will add more heat to the system which move equilibrium in the forward reaction with decrease in temperature. Hence, the equilibrium will shift in the right direction.

So, the concatenation of NO will increase.

7 0
3 years ago
Find the number of molecules in 27 grams of water
Aleks04 [339]

In this question we have given the gram of water and we know that 1 mole of water = 18 gram of water and 27 g of water contain 1.5 g of water 27 / 18 = 1.5 g

As we know that avogandro'S no is equal 6.022*1023

1.5g * 6.022*1023 = 9.0 * 1023 molecules present in each 27 g of water.

I hope you will understand better now if you like then comment below and tell me. Best of luck.

5 0
3 years ago
Based on this chemical equation KNO3 --> KNO2 + O2
TEA [102]
It’s decomposition. The compound is being broken apart into two different things aka decomposing.
6 0
3 years ago
When the following equation is balanced, what is the coefficient for CaCl2 ?
USPshnik [31]
To know if an equation is balanced you need to check and see how much of each molecule is on either side of the arrow. Right now you have 1-Ca, 2-H, 2-Cl on the left side of the arrow and 1-Ca, 2-Cl, and 2-H on the right side too. Because all the molecules are equal on both sides this means that the equation is balanced. So in front of the CaCl2 there is an assumed coefficient of 1. The answer is 1. 
6 0
2 years ago
In a reaction, 24.9 L of N2 reacts with excess Hy to produce NH3. How many liters of NH3
Amiraneli [1.4K]

Answer:

A. 49.8L of NH3.

B. 33.83g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

N2 + 3H2 —> 2NH3

A. From the balanced equation above,

1L of N2 produced 2L of NH3.

Therefore, 24.9L of N2 will produce = 24.9 x 2 = 49.8L of NH3.

Therefore, 49.8L of NH3 is produced from the reaction.

B. Determination of the mass NH3 produced.

First, we shall determine the number of mole of NH3 produced. This can be obtained as follow:

Volume (V) = 49.8L

Pressure (P) = 97.8 kPa = 97.8/101.325 = 0.97atm

Temperature (T) = 23.7°C = 23.7°C + 273 = 296.7K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) =...?

PV = nRT

n = PV /RT

n = (0.97 x 49.8) / (0.082 x 296.3)

n = 1.99 mole

Next, we shall convert 1.99 mole to NH3 to grams. This is illustrated below:

Number of mole NH3 = 1.99 mole

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 1.99 x 17

Mass of NH3 = 33.83g

Therefore, 33.83g of NH3 is produced.

6 0
3 years ago
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