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3 years ago

How many grams of NaOH are produced from 20 grams of Na2Co3

Chemistry

1 answer:

Mariulka [41]3 years ago

4 0

Answer:

1400

Explanation:

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3 years ago

Why is it called alkaline battery?

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3 years ago

Which of the following showing the Law of Conservation of Mass?

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4 years ago

Read 2 more answers

Based on the diagram below, how much of the excess reactant is left over? *

Alexxandr [17]

Answer:

3 of lunchmeat and 2 slices of cheese

Explanation:

From the question given,

Each sandwich contains:

2 bread + 3 lunch meat + 1 cheese.

Now, the limiting reactant is the slice of bread.

We can determine the leftover as follow:

1. For the lunchmeat.

From the simple equation above,

2 bread requires 3 lunchmeat.

Therefore, 6 bread will require = (6 x 3)/2 = 9 lunchmeat.

Lunchmeat given = 12

Lunchmeat required = 9

Leftover lunchmeat = 12 – 9 = 3

Therefore, 3 lunchmeat is leftover.

2. For cheese.

From the simple equation above,

2 bread requires 1 cheese.

Therefore, 6 bread will require = 6/2 = 3 cheese.

Cheese given = 5

Cheese required = 3

Leftover cheese = 5 – 3 = 2

Therefore, 2 cheese is leftover.

From the simple illustrations above,

3 lunchmeat and 2 cheese are leftover.

7 0

3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

3 years ago