Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>
Answer: Antoine Lavoisier.
He classified the elements into four groups: elastic fluids, nonmetals, metals and earths. Some of the called elementes were not really elements (light and heat). Others were compounds, e.g. hydrochloric acid.
<u>Answer:</u> The mass of solid NaOH required is 80 g
<u>Explanation:</u>
Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:

where,
n = acidity for bases = 1 (For NaOH)
Molar mass of NaOH = 40 g/mol
Putting values in above equation, we get:

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.
Mathematically,

Or,
......(1)
We are given:
Given mass of NaOH = ?
Equivalent mass of NaOH = 40 g/eq
Volume of solution = 400 mL
Normality of solution = 5 eq/L
Putting values in equation 1, we get:

Hence, the mass of solid NaOH required is 80 g