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>
<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %
<u>Explanation:</u>
We are given:
A chemical compound having chemical formula of 
It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms
To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:
Mass of compound = ![[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol](https://tex.z-dn.net/?f=%5B%281%5Ctimes%2014%29%2B%285%5Ctimes%201%29%2B%281%5Ctimes%2012%29%2B%283%5Ctimes%2016%29%5D%3D79g%2Fmol)
Mass of hydrogen = 
Putting values in above equation, we get:
Hence, the percent composition by mass of hydrogen in given compound is 6.33 %
C.... 376 tbh just took the test and was right
Answer:
A free body diagram is used to calculate static and dynamic forces acting on an object. In other words, a free body diagram is the starting point to develop a mathematical model to find and calculate various forces acting on a body.
Explanation:
Ok they are both can formed by the same high pressure system . the heat waves are extended periods of above-normal temperatures . but the cold wave is an extended periods of below normal temperatures
