Answer:
14. B 15. D 16. C 17. B
Explanation:
The spontaneous reaction that occurs when the cell operates is shown below:
⇒
We need to select the correct option from the list below for the following questions.
(A) Voltage increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information.
14. A 50-milliliter sample of a 2-molar 
solution is added to the left beaker.
If a 50-milliliter sample of a 2-molar solution is added to the left beaker, the voltage decreases but its value remains greater than zero. The correct option is B
15. The silver electrode is made larger.
If the silver electrode is made larger, no change in the value of the voltage since we don't have the idea of the initial value. The correct option is D.
16. The salt bridge is replaced by a platinum wire.
If the salt bridge is replaced by a platinum wire, there will be no passage of electrons because electrons can't pass through a platinum wire. Therefore, the voltage will be zero and remains at zero. The correct option is C.
17. Current is allowed to flow for 5 minutes.
If current is allowed to flow for 5 minutes, the voltage decreases but its value remains greater than zero. The correct option is B.
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
Where Blocal = local magnetic field between the two regions of the molecule
Blocal = (1-σ)B0
ΔBlocal = (1-σ1)B0 - (1-σ2)B0 = (σ2 - σ1)B0 = ΔσB0 ≈ ΔδB0 x 10∧-6
= (3.36-1.16) x 10∧-6 x B0 = 2.20 x 10∧-6B0
(a) ΔBlocal = 2.20 x 10∧-6 x 1.9T = 4.2 μT
(b) ΔBlocal = 2.20 x 10∧-6 x 16.5T = 36.3 μT
Explanation:
I'm pretty sure all 4 are subatomic particles but if i had to guess i'd be Photons
