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2 years ago

An atom of uranium-238 undergoes radioactive decay to form an atom of thorium-234. Which type of nuclear decay has occurred?

Chemistry

1 answer:

WINSTONCH [101]2 years ago

5 0

Answer:-

Alpha decay

Explanation:-

Uranium 238 has atomic number 92 and mass number 238.

Thorium 234 has atomic number 90 and mass number 234.

So, the change in atomic number as uranium 238 disintegrates into thorium234 = 92 – 90 = 2

So, the change in mass number as uranium 238 disintegrates into thorium234= 238 – 234 = 4

We know that when an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.

So when an atom of uranium 238 undergoes radioactive decay to form an atom of thorium-234, alpha decay has occurred.

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If 2.0 moles of A and 3.0 moles of B react according to the hypothetical reaction below, how many moles of the excess reactant w

valina [46]

.5 mol of A will be left over since 1.5 mol of A will be used for every 3 mol of B due to the 2:1 ratio established by the formula.

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3 years ago

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zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol$

The given reaction equation is as follows.

$C + 2H_{2} \rightarrow CH_{4}$

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

$Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol$

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

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3 years ago

Which part of the atom has to stay the same for it to be the same element? number of neutrons

kolezko [41]

For it to be the same element it must contain the same number of protons

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3 years ago

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The colligative molality of an unknown aqueous solution is 1.56 m.

yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- Vapor pressure lowering

ΔP = P° . Xm . i

- Boiling point elevation

ΔT = Kb . m . i

- Freezing point depressión

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

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3 years ago