Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Answer:
1.57 x 10⁷m
Explanation:
Given quantity is;
1.57 x 10¹⁴nm
Now;
1 nm = 10⁻⁹
So, let us convert this given quantity;
1 nm = 10⁻⁹
1.57 x 10¹⁴nm will give 1.57 x 10¹⁴ x 10⁻⁹ = 1.57 x 10⁷m
Answer:
0
Explanation:
Given parameters:
Half-life = 8.08days
Unknown:
What fraction is left unchanged after 16.16days = ?
Solution:
The half - life of a substance is the time taken for the half of a radioactive material to decay to half.
Day 0 Day 8.08 Day 16.16
100% 50% 0% Parent
0% 50% 100% Daughter
After 16.16 days, non of the original sample will remain unchanged.
The answer is (3). The reaction that can occur at the anode is oxidation reaction which will lose electrons. So (1) and (2) are not correct. For (4) Fe3+ can not lose electrons again.
The student was not successful.
Consider the standard reduction potentials.
Li⁺ + e⁻ ⇌ Li; E° = -3.04 V
2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻; E° = -0.83 V
To reduce Li⁺ to Li, the student must apply 3.04 V.
However, it takes only 0.83 V to reduce water to hydrogen.
Thus, the student will get H₂ instead of Li.
