Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
Answer:
Four
Step-by-step explanation:
The <em>superscripts</em> in an electron configuration tell us how many electrons are in a subshell.
If the electron configuration is 1s¹ 2s¹2p², the total number of electrons is
1 + 1 + 2 = 4
The atom contains four electrons.
<em>Note</em>: this atom is in an <em>excited state</em>, because the 1s and 2s subshells can each hold one more electron.
Ni (Nickel) is the best conductor, and in fact the only conductor, in that list
Answer:
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Explanation:
Reduction half reaction
2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-
Oxidation half reaction
2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e
Balanced overall equation
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
