Answer:
D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.
Explanation:
- It is a redox reaction that is consisted of two half-reactions:
Oxidation reaction:
Zn losses 2 electrons and is oxidized to Zn²⁺:
<em>Zn → Zn²⁺ + 2e⁻.</em>
<em></em>
Reduction reaction:
H⁺ gains 1 electron and is reduced to H:
<em>2H⁺ + 2e⁻ → H₂.</em>
<em></em>
<em>So, the right choice is: D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.</em>
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Answer:
Mass of one electron is 9.1 × 10⁻³¹ kg
Mass of one proton is 1.673 × 10⁻²⁷ Kg
Mass of one neutron is 1.675 × 10⁻²⁷ Kg
<u>-TheUnknownScientist</u><u> 72</u>
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
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)
As you move from left to right across a period,the number of valence electrons<span>-increases. </span>
