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2 years ago

Which of the following is not a balanced nuclear equation?

Chemistry

1 answer:

charle [14.2K]2 years ago

6 0

Answer:

¹⁰₄Be —> ¹⁰₃Li + ⁰₋₁β

Explanation:

To know which equation is unbalanced, we shall compare the mass number and atomic number of the reactant and product together. This is illustrated below:

¹²¹₅₀Sn —> ⁰₋₁β + ¹²¹₅₁Sb

121 = 0 + 121

121 = 121

50 = –1 + 51

50 = 51

Thus the above equation is balanced

¹⁹⁰₇₈Pt —> ¹⁸⁶₇₆Os + ⁴₂α

190 = 186 + 4

190 = 190

78 = 76 + 2

78 = 78

Thus, the above equation is balanced

¹⁰₄Be —> ¹⁰₃Li + ⁰₋₁β

10 = 10 + 0

10 = 10

4 = 3 + (–1)

4 = 3 – 1

4 = 2

Thus, the above equation is unbalanced.

²³⁸₉₂U —> ⁴₂α + ²³⁴₉₀Th

238 = 4 + 234

238 = 238

92 = 2 + 90

92 = 92

Thus, the above equation is balanced.

From the illustrations made above, only ¹⁰₄Be —> ¹⁰₃Li + ⁰₋₁β

Is unbalanced.

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A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the

Stells [14]

Answer:

e. T₂= 4T₁

Explanation:

Initially, we have a number of moles (n₁) a gas sample at a certain pressure (P), temperature (T₁) and volume (V). We can relate these variables through the ideal gas equation.

P . V = n₁ . R . T₁

where,

R is the ideal gas constant

We can rearrange this equation like:

$T_{1}=\frac{P.V}{n_{1}.R}$

If only one fourth of the initial molecules remain n₂ = 1/4 n₁. The new temperature (T₂) assuming pressure and temperature remain constant is:

$P.V=n_{2}.R.T_{2}=\frac{1}{4} n_{1}.R.T_{2}\\\frac{P.V}{n_{1}.R} =\frac{1}{4} T_{2}\\T_{1}=\frac{1}{4} T_{2}\\T_{2}=4.T_{1}$

6 0

3 years ago

5.75 moles of H2 gas to liters at STP

antiseptic1488 [7]

5.75mol (22.4L/1mol) = 128.8L

8 0

2 years ago

Name of the the following unsaturated Hydrocarbon

djyliett [7]

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2 years ago

What type of reaction produces a metal from its metal oxide

Nataly [62]

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2 years ago

Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance

ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0

3 years ago