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3 years ago

Which represents the balanced equation for the beta minus emission of phosphorus-32?

Chemistry

2 answers:

Agata [3.3K]3 years ago

8 0

Answer: The correct option is C.

Explanation: Beta minus ( $\beta ^-$ ) decay is the decay in which a neutron is converted into proton and releases beta particle or generally known as electron.

There is an increase in the Atomic number because one protons is being formed.

In the case of beta-minus decay of phosphorous, the P-nucleus gets changed to Sulfur nucleus. The balanced Chemical equation for this process is:

$^{32}_{15}\textrm{P}\rightarrow ^{32}_{16}\textrm{S}+^0_{-1}\beta$

Hence, the correct option is C.

uranmaximum [27]3 years ago

3 0

Answer : Option C) $P^{32}_{15} ---> S^{32}_{16} +e^ {0}_{-1}$

Explanation : The balanced equation for the beta minus decay for emission of Phosphorus-32 is $P^{32}_{15} ---> S^{32}_{16} +e^ {0}_{-1}$

Where, it is clearly seen that in the beta minus emission a neutron decays into a proton, an electron, and an anti-neutrino.

So, in the phosphorus-32 beta minus decay the product is sulfur-32 along with an electron created from the anti-neutrino.

Therefore, option C is the correct option.

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2 years ago

How many milliliters of a 0.266 m lino3 solution are required to make 150.0 ml of 0.075 m lino3 solution?

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We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,

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3 years ago

Read 2 more answers

Using the reagents below, list in order (by letter, no period) those necessary to transform 1-chlorobutane into 1-butyne. Note:

Andru [333]

Answer:

gde

Explanation:

We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.

Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.

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3 years ago

An experiment reveals that 125.0 grams of an unknown metal increases in temperature from 22.0 oC to 43.6 oC upon absorbing 640 j

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Answer:

Cp = 0.237 J.g⁻¹.°C⁻¹

Explanation:

Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.

The equation used for this problem is as follow,

Q = m Cp ΔT ----- (1)

Where;

Q = Heat = 640 J

m = mass = 125 g

Cp = Specific Heat Capacity = <u>??</u>

ΔT = Change in Temperature = 43.6 °C - 22 °C = 21.6 °C

Solving eq. 1 for Cp,

Cp = Q / m ΔT

Putting values,

Cp = 640 J / (125 g × 21.6 °C)

Cp = 0.237 J.g⁻¹.°C⁻¹

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