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There are a lot
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>
Explanation:
It is given that,
The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :



So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.
100 by 500 ml so if you put 50 ml in 630torr it would evaporate
Explanation:
it is more than ¹00 celcwius so it would eveporate
Answer:

Explanation:
= Initial volume = 100 mL
= Final volume = 1000 mL
= Initial concentration = 0.5 M
= Final concentration
We have the relation

The new concentration is
.