Answer:
0.25 L of a solution with a molarity of 6M has 6*0.25 = 1.5 moles of the solute. The molar mass of ammonium sulfate is 132.14 g/mole. The mass of 1.5 moles is 132.14*1.5 = 198.21 g. Therefore 198.12 g of ammonium sulfate are required to make 0.25 L of a solution with a concentration of 6M.
Explanation:
Answer:
333K
Explanation:
Answer is actually 333.15, but to round it off, is 333
Answer: A, 1.3*1020, 1326
A. 1326
B. 2960.9
C. 9804
D. 8559.6
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Atomic number and the number of protons are the same...
Neutrons = Mass number - number of protons
Electrons are same # unless there is a charge
The whole number you see on the periodic table is the atomic number of the element which is also same as the number of protons
1) carbon - 14 ; Mass number = 14 , Protons = 6 , Neutrons = 14 - 6 = 8
Electrons = 6
2) Lead - 208 ; Mass # = 208 , Protons = 82 , Neutrons = 208 - 82 = 126
Electrons = 82
3) Uranium - 239 ; Mass # = 239 , Protons = 92,Neutrons = 239 - 92 = 147
Electrons = 92
4) Uranium - 238 ; Mass # = 238 , Protons = 92 , Neutrons = 238 - 92 = 146
Electrons = 92
5) Tin - 118 ; Mass # = 118 , Protons = 50 , Neutrons = 118 - 50 = 68
Electrons = 50
