Answer:
hydrogen ions
Explanation:
because acid is the specie that have ability to donate proton or forming bond with electron pair
The answer to the question is b
3. 4 g of a nonelectrolyte dissolved in 78. 3 g of water produces a solution. The molar mass of the solute will be 17.94.
<h3>
What is molar mass?</h3>
Molar mass of a substance is its mass in grams in per mole of a solution.
Freezing point: Freezing point of a substance is a temperature at which a liquid starts to solidify.
Depression in the freezing point can be calculated
[Depression in freezing point of pure solvent—Freezing point of solution] =[(0) - (-4.5)] °C =4.5 °C
molar mass = Number of moles of solute m / Mass of solvent in Kg
3.4g / M x 1/ 0.0783 kg = 43.42
Substitute AT by 4.5°C , Kr by 1.86 °C/m, and m by 43.42 m in equation (1) as follows:
1.86 x 43.42 / 4.5 = 17.94
Therefore, molar mass of solute to be 17.94.
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Answer:
The answer to your question is 0.113 moles of Fe₂O₃
Explanation:
Data
moles of Fe₂O₃ = ?
mass of Fe₂O₃ = 18 grams
Process
1.- Calculate the molar mass of Fe₂O₃
Fe₂O₃ = (56 x 2) + (16 x 3)
= 112 + 48
= 160 g
2.- Use proportions to solve this problem. The molar mass is equivalent to 1 mol.
160 g of Fe₂O₃ --------------- 1 mol
18 g of Fe₂O₃ ---------------- x
x = (18 x 1)/160
x = 0.113 moles of Fe₂O₃
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be 
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio 
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
