How many kilograms are in 4 moles of Na2CO3?

1 answer:

6 0

First you need to find the amount of mass of Na2CO3 in one moles
(Use periodic chart)

Na= 22.99 x 2 = 45.98
C = 12.01
O = 16.00 x 4 = 64.00

Add the molar masses together to get 121.99

To find how many grams are in 4 moles, times 121.99 by 4
This gives you 487.96

But the questions asks for the answer to be in kilograms nor grams, to change into kilograms divide by 1000
This gets you the answer: 0.49 kg

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Determine whether a precipitate will form when 0.96g of Na2CO3 is combined with 0.20g BaBr2 in a 10 L solution.

Mashcka [7]

Answer:

Qsp > Ksp, BaCO3 will precipitate

Explanation:

The equation of the reaction is;

Na2CO3 + BaBr2 -------> 2NaBr + BaCO3

Since BaCO3 may form a precipitate we can determine the Qsp of the system.

Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles

concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M

Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles

concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M

Hence;

[Ba^2+] = 6.7 * 10^-5 M

[CO3^2-] = 9.1 * 10^-4 M

Qsp = [6.7 * 10^-5] [9.1 * 10^-4]

Qsp = 6.1 * 10^-8

But, Ksp for BaCO3 is 5.1*10^-9.

Since Qsp > Ksp, BaCO3 will precipitate

3 0

3 years ago

What is Y? 222/86 Rn A/z + 4/2He

kondaur [170]

$^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}$

Y is Po-218.

A = 218
Z = 84.

<h3>Explanation </h3>

$^{222}_{\phantom{2}86}\text{Rn} \to ^{A}_{Z}\text{Y}+ ^{4}_{2}\text{He}$

Here's the symbol of a particle in a nuclear reaction. $^{A}_{Z}\text{Y}$ .

A stands for mass number. Z stands for atomic number. Both numbers shall conserve in a nuclear reaction.

The mass number on the left hand side is 222.
The two mass numbers on the right hand side add up to A + 4.
222 = A + 4.
A = 218

So is the case for the atomic number. Try figure out the atomic number of Y using the same approach.

The atomic number on the left hand side is 86.
The two atomic number on the right hand side add up to __ + __.
86 = __ + __.
Z = 84.

What element is Y? The atomic number of Y is 84. Refer to a periodic table. Element 84 corresponds to Po (polonium). Y is Polonium-218. The symbol of Y should be written as $^{218}_{\phantom{2}84}\text{Po}$ . Hence the equation: