The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
I think the correct answer would be C. The expression that would best represent a second order rate law would be r =k[X][Y]. Reaction with this rate law are those that depend on the concentration of two first order reactants or a second order reactant.
The prefix milli means thousand so the correct conversion factor is 1000mg/g
Answer:
substance
Explanation:
A mixture is when two or more <u>different</u> atoms/molecules are together, but not joined.
A substance is when the <u>same </u>atom/molecule is in a group together.
In this example, it is a substance because it is comprised of the same molecule not joined all together. If you wanted a mixture, other colored atoms/molecule (e.g. add green atoms) would change it to this property.
In No3-1 the oxidation number of oxygen is -5 so oxidation number of N would be +5
