Answer:
Explanation:
The Ce metal has electronic configuration as follows
[Xe] 4f¹5d¹6s²
After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.
Eu has electronic configuration as follows
[ Xe ] 4 f ⁷6s²
[ Xe ] 4 f ⁷
Its outermost orbit contains 2 electrons so Eu²⁺ is stable. Its +3 oxidation state is also stable.
Ce⁺²
Ionic bonds are a metal and a non metal bond and a covalent bond is two no metals banded together.
Answer is:
7.8 lb of 21% aluminum and 33.2 ib of <span>
42% aluminum.</span>
ω₁<span> = 21% ÷ 100% = 0.21.
ω</span>₂<span> = 42% ÷ 100% = 0.42.
ω</span>₃<span> = 38% ÷ 100% = 0.38.
</span>m₁ = ?.
m₂<span> = ?.
</span>m₃ = m₁ + m₂<span>.
</span>m₃ = 41 pounds.
m₁ = 41 lb - m₂<span>.
ω</span>₁ · m₁ + ω₂ ·m₂ = ω₃ · m₃.
0.21 · (41 lb -
m₂) + 0.42 · m₂ = 0.38 · 41 lb.
8.61 lb - 0.21m₂ + 0.42m₂ = 15.58 lb.
0.21m₂ = 6.97 lb.
m₂ = 6.97 lb ÷ 0.21.
m₂ = 33.2 lb.
m₁ = 41 lb - 33.2 lb.
m₁<span> = 7.8 lb.</span>
The answer is a change in internal energy causes work to be done and heat to flow into the system.
<u>Explanation:</u>
Boyle's law says, PV=RT
- Here P represents the pressure, V represents the volume and T represents the temperature. R is a constant. The volume of an ideal gas is inversely proportional to its pressure if the temperature is constant.
- When a bubble is present in deep water it has water pressure and atmospheric pressure. Then the Volume increases when water pressure raises which is proportional to the depth reduces.
- But we should not finalize the volume of the bubble will be four-time as great as at the top than the bottom. if the bottom of the lake is at four atmospheres, the temperature will not be equal to the top.
- If the bubble travels from the bottom to the top or vice-versa, it's going to lose or gain heat in a way that must be quite hard to measure.
Answer:
Explanation:
Given
The data in the table
Required
Follow the steps appended to the question;
Step 1: Calculate the Mean or Average
Mean = Summation of lengths divided by number of teams;
Step 2: Get The Range
Step 3: Divide Range by 2
Step 4: Determine the Precision
Substitute 2.70 for Average and 0.1 for Approximate Range
