Explanation:
Force of attraction is larger in neon -60°C
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
The temperature would increase as there would be more global warming
Answer:
D. Ni²⁺
Explanation:
We know at once that the answer cannot be A or C, because Ni and Cu are already in their lowest oxidation states.
The correct answer must be either B or D.
An electrolytic cell is the opposite of a galvanic cell. In the former, the reaction proceeds spontaneously. In the latter, you must force the reaction to occur.
One strategy to solve this problem is:
- Look up the standard reduction potentials for the half reaction·
- Figure out the spontaneous direction.
- Write the equation in the reverse direction.
1. Standard reduction potentials
E°/V
Cu²⁺ + 2e⁻ ⟶ Cu; 0.3419
Ni²⁺ + 2e⁻ ⟶ Ni; -0.257
2. Galvanic Cell
We reverse the direction of the more negative half cell and add.
<u>E°/V
</u>
Ni ⟶ Ni²⁺ + 2e⁻; 0.257
<u>Cu²⁺ + 2e⁻ ⟶ Cu; </u> 0.3419
Ni + Cu²⁺ ⟶ Cu + Ni²⁺; 0.599
This is the spontaneous direction.
Cu²⁺ is reduced to Cu.
3. Electrochemical cell
<u>E°/V</u>
Ni²⁺ + 2e⁻ ⟶ Ni; -0.257
<u>Cu ⟶ Cu²⁺ + 2e⁻; </u> <u>-0.3419</u>
Cu + Ni²⁺ ⟶ Ni + Cu²⁺; -0.599
This is the non-spontaneous direction.
Ni²⁺ is reduced to Ni in the electrolytic cell.
Answer:
T = 3206.89 K
Explanation:
Given that,
Number of moles, n = 1.9 moles
Pressure, P = 5 atm
Volume, V = 100 L
We need to find the temperature of the gas. Let the temperature is T. We know that,
PV = nRT
Where
R is gas constant, R = 0.08206 L-atm/mol-K
Put all the values,
So, the temperature of the gas is 3206.89 K.
