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3 years ago

What is the molar mass of calcium bromide?

Chemistry

1 answer:

goblinko [34]3 years ago

7 0

CaBr2 = 40.078 + 79.9 x 2 => 199.878 g/mol

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Which of the following is true for the actual yield of a reaction?

ludmilkaskok [199]

Answer:

It is always less than the theoretical yield

Explanation:

For many chemical reactions, the actual yield is usually less than the theoretical yield. This is due to possible loss in the process or inefficiency of the chemical reaction.

4 0

3 years ago

Read 2 more answers

Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t

Veseljchak [2.6K]

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

$4NH_3+7O_2\rightarrow 4NO_2+6H_2O$

The expression for rate of reaction for the reactant :

$\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}$

$\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}$

The expression for rate of reaction for the product :

$\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}$

From this we conclude that, all the options are correct.

3 0

3 years ago

Convert 2.17 x 10 -8 Mg into pg

viktelen [127]

This answer is 24 because 2.17 x 10 -8 is 24 so that would be your answer

4 0

3 years ago

Analysis of an ore of calcium shows that it contains 13.61 g calcium and 21.77 g oxygen in a 46.28-g sample. Calculate the perce

kondor19780726 [428]

Answer:

29.41% of Calcium and 47.04% of Oxygen

Explanation:

The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.

The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:

13.61g / 46.28g * 100 = 29.41% of Calcium

And percent composition of Oxygen is:

21.77g / 46.28g * 100 = 47.04% of Oxygen

5 0

3 years ago

What is the volume, in liters, of 1.40 mol of oxygen gas at 20.0°C and 0.974 atm?

topjm [15]

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, $R=0.08206\ L-atm/mol-K$

Finding for V,

$V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L$

So, the volume of the gas is 34.55 L.

4 0

3 years ago