Answer:
60.0mL of the diluted solution are needed
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.
Explanation:
As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:
3 * 20.0mL = 60.0mL of the diluted solution are needed
Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:
1.00M / 0.100M = 10 times must be diluted the solution.
As we need at least 60.0mL, the minimum volume of the stock solution must be:
60.0mL / 10 times =
<h3>6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.</h3>
Answer:
40 mL
Explanation:
V2=P1V1/P2
You can check this by knowing that P and V at constant T have an inverse relationship. Hence, this is correct.
The compound is made up of multiple atoms.
Answer:
The value of Kc is 660 (Option E) is correct
Explanation:
Step 1: Data given
Kp = 27
Temperature = 25.0 °C
Step 2: The balanced equation
2 NO(g) + Br2(g) ⇄ 2 NOBr(g)
Step 3: Calculate Kc
Kp = Kc * (RT)^Δn
⇒ with Kp = 27
⇒ with Kc = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = The temperature = 25 °C = 298 K
⇒ Δn = the difference in moles = -1
27 = Kc * (0.08206*298)^-1
Kc = 660
The value of Kc is 660 (Option E) is correct
Answer:
C. copper.
Explanation:
- The atom which loses electrons (its oxidation sate be more positive) is the atom that is oxidized.
- While, the atom which gains electrons (its oxidation sate be more negative) is the atom that is reduced.
It is oxidation sate is changed from (+5) in the reactants (NO₃⁻) to (+4) in the products (NO₂). N gains 1 electron
So, it is reduced.
It is oxidation sate is the same (-2) in the reactants (NO₃⁻) and (-2) in the products (NO₂).
<em>So, it is neither be oxidized nor reduced.</em>
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It is oxidation sate is changed (0) in the reactants (Cu) to (+2) in the products (Cu²⁺). Cu loses 2 electrons.
<em>So, it is oxidized.</em>
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It is oxidation sate is the same (+1) in the reactants (H⁺) and (+1) in the products (H₂O).
<em>So, it is neither be oxidized nor reduced.</em>
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