The elements in this list are mercury, gold, iron, carbon and hydrogen. The compounds in this list, on the other hand, are sucrose, table salt, water and air. Elements are composed only of one substance while compounds are composed of two or more substances.
The answer is 60.3% magnesium, 39.7% oxygen.
Solution:
The chemical equation for the reaction is 2 Mg + O2 → 2 MgO.
Since magnesium reacts completely with oxygen, it is the limiting reactant in the reaction. Hence, we can use the number of moles of magnesium to get the mass of MgO produced:
moles of magnesium = 14.7g / 24.305g mol-1
= 0.6048 mol
mass of MgO = 0.6048mol Mg(2 mol MgO/2mol Mg)(40.3044g MgO/1 mol MgO)
= 24.376g MgO
We can now solve for the percentage of magnesium:
% Mg = (14.7g Mg / 24.376g MgO)*100% = 60.3%
We also use the number of moles of magnesium to get the mass of oxygen consumed in the reaction:
mass of O2 = 0.6048 mol Mg (1mol O2 / 2mol Mg) (31.998g / 1mol O2)
= 9.676g
The percentage of oxygen is therefore
% O2 = (9.676g O2 / 24.376g MgO)*100%
= 39.7%
Notice that we can just subtract the magnesium's percentage from 100% to get
% O2 = 100% - 60.3% = 39.7%
Answer:
5.83 g
Explanation:
First, you must start with a balanced equation so you can see the mole ratios.
NaOH + H₃BO₃ --> NaBO₂ + 2H₂O
You can see that it takes 1 mole of sodium hydroxide to form 1 mole of sodium borate. 1:1 ratio
Now you must calculate how many moles of NaOH 35.47 g equals.
Na = 22.99 amu
O = 15.99 amu
H = 1.008 amu
NaOH = 39.997 amu
35.47 g ÷ 39.997 amu = 0.08868 moles of NaOH
Since it's a 1:1 ratio, the same number of moles of NaBO₂ is created. Now you must convert moles to grams.
Na = 22.9 amu
B = 10.81 amu
2 O = 31.998 amu
NaBO₂ = 65.798 amu
0.08868 moles x 65.798 = 5.83 g
Answer:
The volumetric ratio is 0,71
Explanation:
Let's begin with the equation:
(1)
Where:
Db: Blend Density, Mb: Blend Mass and Vb: Blend Volume
And we know: 
(2)
Where:
Vg: Gasoline Volume and Vk: Kerosene Volume
Therefore replacing (2) into (1):
(3)
Where:
Dg: Gasoline Density and Dk: Kerosene Density
The specific gravity is defined as:
Therefore:
Where:
Dref: Reference Density
SGb: Blend Specific Gravity
SGg: Gasoline Specific Gravity (which is 0.7 approximately)
SGk: Kerosene Specific Gravity
Replacing these equations into (3) we get:
Replacing with the Specific Gravity data, we obtain:
It is an element. Aluminun foil is aluminum prepared in thin leaves.
