<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
Answer:
Explanation:
We have to start with the <u>reaction</u>:

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:
C: 12 g/mol
O: 16 g/mol
Mg: 23.3 g/mol
If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:
In other words:
. With this in mind, we can calculate the moles:

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

With the molar mass of
(
. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:
I hope it helps!
Answer:
no .........................
A
The number of electrons/protons is the one that determines the atomic number of an element.
Answer:
- <u>You need to convert the number of atoms of Ca into mass in grams, using Avogadro's number and the atomic mass of Ca.</u>
Explanation:
The amount of matter is measured in grams. Thus, you need to convert the number of atoms of Ca (calcium) into mass to compare with 2.45 grams of Mg.
To convert the atoms of calcium into mass, you divide by Avogadro's number, to obtain the number of moles of atoms, and then divide by the atomic mass of calcium.
<u />
<u>1. Number of moles, n</u>

<u />
<u>2. Mass</u>
- mass = number of moles × atomic mass
- mass = 0.053969mol × 40.078g/mol = 2.16g
Then, 2.45 g of Mg represent a greaer mass than the 3.25 × 10²² atoms of Ca.