Answer:
10^-14 / [H3O+]
Explanation:
Hydronium (H3O+) and hydroxide ions (OH-) are both present in pure water and in all aqueous solutions.
Their respective concentrations in water are 10^-7 M each and are inversely proportional to each other as given by the ion product of water, Kw.
Kw = [H3O+][OH−]
Where Kw = 1.0 * 10^-14,
[H3O+] = concentration of hydronium ions
[OH-] = concentration of hydroxide ions
Therefore, [OH-] = Kw / [H3O+]
[OH-] = 10^-14/[H3O+]
Answer: C6H6
Explanation: Molecular Formula is the molecular mass divided by the empirical formula mass of CH which is C 12and H is 1 so 12 + 1= 13 g CH
Solution:
78.11 g CH / 13 g CH = 6
Answer:
A
Explanation:
By nature, a hybrid has one dominante allele and one recessive allele
D. There is one atom of sulfur and three atoms of oxygen.
The "tri" in trioxide indicates three so there are three atoms of oxygen. As nothing is said about the sulfur, there is just one atom.
Hope this helps!!
The maximum amount of hydrogen gas that can be prepared is if all the hydrogen from both compounds is released.
The hydrogen in 4.94 g of SrH2 is calculated from the mass ratios between Sr and H
1) H2 in SrH2
Sr atomic mass = 87.62 g/mol
H2 molar mass = 2.02 g/mol
Mass of 1 mol of SrH2 = 87.62 g / mol + 2.02 g/mol = 89.64 g/mol
Ratio of H2 to SrH2 = 2.02 g H2 / 89.64 g SrH2
Proportion: 2.02 g H2 / 89.64 gSrH2 = x / 4.93 g SrH2
=> x = 4.93 g SrH2 * 2.02 g H2 / 89.64 g SrH2 = 0.111 g H2
2) H2 in H2O
2.02 g H2 / 18.02 g H2O * 4.14 g H2O = 0.464 g H2
3) Total mass of hydrogen = 0.111 g + 0.464 g = 0.575 g
Answer: 0.575 g
