1. O2 is not a compound because it only contains one or more type of the same element atom.
2. O2 is a molecule because a molecule is one or more of the same element atom.
3. The law of conversion is that the mass of the system will stay the same when transfer takes place. Like if you had an equation O+H2—> H2O the mass will remain the same.
4. It will be equal to 10 because of law of conservation of matter.
5. One observation can be that the compound, reaction you’re observing, has change states.
Answer:
9.1 mol
Explanation:
The balanced chemical equation of the reaction is:
CO (g) + 2H2 (g) → CH3OH (l)
According to the above balanced equation, 2 moles of hydrogen gas (H2) are needed to produce 1 mole of methanol (CH3OH).
To convert 36.7 g of hydrogen gas to moles, we use the formula;
mole = mass/molar mass
Molar mass of H2 = 2.02g/mol
mole = 36.7/2.02
mole = 18.17mol
This means that if;
2 moles of H2 reacts to produce 1 mole of CH3OH
18.17mol of H2 will react to produce;
18.17 × 1 / 2
= 18.17/2
= 9.085
Approximately to 1 d.p = 9.1 mol of methanol (CH3OH).
Answer:
8.99×10^-7m
Explanation:
The wavelength can be calculated using the expression below
E=hcλ
Where E= energy= 2.21 x 10^-19 J.
C= speed of light= 3x10^8 m/s
h= planks constant= 6.626 × 10^-34 m2 kg / s
E=hcλ
λ= E/(hc)
Substitute for the values
λ=( 2.21 x 10^-19 )/(6.626 × 10^-34 × 3x10^8 )
= 8.99×10^-7m
Answer:
87.15%
Explanation:
To find percent yield, we can use this simple equation
Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.
They give us these values, so to find the percent yield, just plug the numbers in.
So, the percent yield is 87.15%
An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.
In NaMnO₄, Mn has the highest oxidation number.
The question is incomplete, the complete question is;
Which of the following species contains manganese with the highest oxidation number?
A) Mn
B) MnF₂
C) Mn₃(PO₄)₂
D) MnCl₄
E) NaMnO₄
In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.
1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.
2) For MnF₂;
Mn has an oxidation number of +2
3) For Mn₃(PO₄)₂
Mn has an oxidation number of +2
4) For MnCl₄
Mn has an oxidation number of +4
5) For NaMnO₄
Mn has an oxidation number of +7
Hence in NaMnO₄, Mn has the highest oxidation number.
Learn more: brainly.com/question/10079361
