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3 years ago

An ion of sulfur has a smaller atomic radius than a neutral atom of sulfur because it loses an electron when becoming an ion.

1 answer:

4 0

Answer is FALSE. Because sulfur has no tendency to loss an electron, since its electronic configuration is 16, it always want to gain 2 electron to form S^2-, following octet rule.

The answer would be yes if it can lose an electron(because when an atom loses an electron it form cation whose nuclear charge remain as same as neutral atom but number of electron decreases. As a result the nuclear charge has to act on a fewer number of electron and hence there will be more attraction between the nuclear charge and outermost electrons resulting in decrease in size) but it is not possible in case of sulfur.

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How does nuclear fission of Uranium - 234 result in electricity being generated?

nevsk [136]

Answer:

See explanation

Explanation:

The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.

As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.

7 0

3 years ago

How many microliters are in 7.4 x 10^-61 centimeters?

schepotkina [342]

Answer:

Microliters = $7.4\times 10^{-58}$

Explanation:

SI unit of volume is liter.

$1 dm^{3}=1 liter$ ..............................(1)

1dm=10cm

$1dm^{3} = 1000 cm^{3}$ .................(2)

replacing value of $1 dm^{3}$ from equation (1) in equation (2)

$1 liter = 1000 cm^{3}$ ...................(3)

but

$1 liter = 10^{6} microliters$ ....................(4)

replacing value of 1 liter from equation (3) in equation (4)

$1000 cm^{3} = 10^{6} microliters$

on solving further ,we get

$1 cm^{3} = \frac{10^{6}}{10^{3}}microliter$

$1 cm^{3} = 1000 microliter$

$7.4\times 10^{-61}\times 1000$

Microliters = $7.4\times 10^{-58}$

8 0

3 years ago

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

7 0

3 years ago

What percentage of strontium-90 remains in a sample after three half-lives have passed? 12.5% 15% 25% 60%?

Lyrx [107]

12.5% of strontium-90 would remain in a sample after three half-lives have passed. Half-life automatically means 50% of the original amount would remain.

4 0

3 years ago

Read 2 more answers

Glaciation in Alaska is a result of interactions between which of the following "spheres"? Check all that apply.

elena-s [515]

I would say E is the right answer.

4 0

4 years ago

Read 2 more answers