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:
1.70 g.cm⁻³
Solution:
Data Given;
Mass = 84.7 g
Volume = 49.6 cm³
Density = ?
Formula Used;
Density = Mass ÷ Volume
Putting values,
Density = 84.7 g ÷ 49.6 cm³
Density = 1.70 g.cm⁻³
Answer:
The value of the partial pressure of the oxygen 
= 690 torr
Explanation:
Total pressure of the mixture of gases = 736 torr
The partial pressure of water vapor = 46 torr
From the law of pressure we know that
Total pressure = The partial pressure of water vapor + The partial pressure of oxygen 
Put the values of pressures in above equation we get,
⇒ 736 = 46 +
⇒ = 736 - 46
⇒ = 690 torr
This is the value of the partial pressure of the oxygen.
Answer:
n=N/Na
n = \frac{8.23 \times {10}^{22} }{6.02 \times {10}^{23} } = 0.1367 \: mol
answer: 0.14 mol
Explanation:
