We are given an equation 2Mg+O2-> 2MgO and a starting chemical Mg of about 212 g. In order to solve for the amount of O2 needed, we need the molecular weight of Mg and O2.
Molecular weight:
Mg=24.305 g/mol
O2=16(2)=32 g/mol
Note that for every 1 mol of O2, the amount of Mg must be 2 mol.
So,
g O2 = 212 g Mg x1mo Mgl/24.305 g Mg x1mol O2 /2 mol Mg x 32 g O2/mol O2
gO2=139.56 g
Therefore, 139.56 g of O2 is needed for every 212 g Mg.
First, we have to correct the equation in the question to b(g)⇆ 1/2 A(g)
at the first equation A(g)⇆ 2 B(g) so,
Kc = [B]^2 [ A] = 0.03
by reverse the equation 2B⇆ A
∴ Kc(original) = [A] / [B]^2
= 1/0.03 = 33 M^-1
and the new equation B⇆ (1/2) A
So, the new Kc = √Kc(original = √33
∴ KC = 5.7
The greatest concentration of atomic mass is the atom's nucleus. This is because the nucleus is made up out of protons and neutrons while the electrons surrounding the nucleus have a very small mass.
Answer:
Hope this helps kinda got confused mid way
Explanation:
So the hydrogen sulphide is H2S for the molecular formula. Then the mass of hydrogen sulphide would be 1×2+32= 34g/mol At STP.
If o100 ml of 3. zero m solution were diluted to 250 ml awareness of<u> 1.2 M.</u>
The concentration of the solution tells you how tons solute has been dissolved within the solvent. for example, in case you upload one teaspoon to two cups of water, the awareness could be stated as 1 t salt in keeping with 2 c water.
V1 = one hundred mL, M1 = 3 M
V2 = 250 mL, M2 = ?
observe Dilution regulation
M1*V1 = M2*V2
so
M2 = M1*V1/V2
M2 = 100/250*3
M2 = 1.2 M
Concentration (normally uncountable, plural concentrations) The act, procedure, or capability of concentrating; the system of becoming focused, or the country of being concentrated. The course of interest to a particular object. The act, technique or product of reducing the quantity of a liquid,
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