Answer:
If you double the mass of an object, you double the kinetic energy. If you double the speed of an object, the kinetic energy increases by four times. The word "kinetic" comes from the Greek word "kinesis" which means motion. Kinetic energy can be passed from one object to another in the form of a collision.
Explanation:
To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.
Mg = 24.305 x 1 = 24.305 amu
O = 15.9994 x 2 =31.9988 amu
H = 1.0079 x 2 = 2.0158 amu
Then, add all the components to get the atomic mass of the molecule.
24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu
The atomic mass is just equivalent to its molar mass.
So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.
Hey there!:
Molar mass of Mg(OH)2 = 58.33 g/mol
number of moles Mg(OH)2 :
moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles
Molar mass of H3PO4 = 97.99 g/mol
number of moles H3PO4:
moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles
Balanced chemical equation is:
3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O
3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !
Now , we will use Mg(OH)2 in further calculation .
Molar mass of Mg3(PO4)2 = 262.87 g/mol
According to balanced equation :
mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2
= (1/3)*0.5246
= 0.1749 moles of Mg3(PO4)2
use :
mass of Mg3(PO4)2 = number of mol * molar mass
= 0.1749 * 262.87
= 46 g of Mg3(PO4)2
Therefore:
% yield = actual mass * 100 / theoretical mass
% = 34.7 * 100 / 46
% = 3470 / 46
= 75.5%
Hope that helps!
Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
The percent composition of this compound :
Mg = 72.182%
N = 27.818%
<h3>Further explanation</h3>
Given
9.03 g Mg
3.48 g N
Required
The percent composition
Solution
Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements
Total mass of the compound :
= 9.03 g + 3.48 g
= 12.51 g
The percent composition :
Mg : 9.03/ 12.51 g x 100% = 72.182%
N : 3.48 / 12.51 g x 100% = 27.818%
