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Alona [7]
4 years ago
14

There are several technological applications for the transuranium elements (Z > 92). An important one is in smoke detectors,

which can use the decay of a tiny amount of americium-241 to neptunium-237.What subatomic particle is emitted from that decay process? Write a balanced nuclear equation for this process.
Chemistry
1 answer:
timama [110]4 years ago
7 0

Answer:

An alpha particle is emitted.

95241Am → 42α + 23793Np

Explanation:

If we look at the difference between the atomic masses of the Americium isotope and neptunium isotope, we see a difference of 4 I e 241 - 237 = 4

Also,if we look at the difference between the atomic numbers of both nuclei, we get a difference of 2 I.e 95 - 93 = 2

Both differences show the exact properties of an alpha decay.

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Iron filings are placed in powdered sulfur. Both iron and sulfur are elements. The iron and sulfur are stirred together until th
Anton [14]
An element I think this is the. Aswer for it
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4 years ago
Explain why you think learning about the rock cycle will help us better understand the world around us.
Karolina [17]

Answer:

You can learn about common things happening to the environment that aren't good. Such as erosion, and the effects it has on the land surrounding it.

Explanation:

3 0
3 years ago
Trichloroethylene, a widely used degreasing solvent for machine parts, is produced in a two-step reaction sequence. Ethylene is
solmaris [256]

Answer:

ΔH°C2H2Cl4(l) = -333,36 kJ/mol

ΔH°r₂ = -35,14 kJ/mol

Explanation:

The ΔH°r of the first reaction is:

ΔH°r = -385,76 kJ/mol = (ΔH°C2H2Cl4(l) + ΔH°H2(g)) - (ΔH°C2H4(g) + 2ΔHCl2 (g))

ΔH°H2(g) = 0 kJ/mol

ΔH°C2H4(g) = 52,4 kJ/mol

Δ°HCl2 (g) = 0 kJ/mol

Replacing:

ΔH°C2H2Cl4(l) = -385,76 kJ/mol + 52,4 kJ/mol = <em>-333,36 kJ/mol</em>

The standard heat of the second reaction is:

ΔH°r₂ = ΔH°C2HCl3(l) + ΔH°HCl(g) - ΔH°C2H2Cl4(l)

Where:

ΔH°C2HCl3(l) = -276,2 kJ/mol; ΔH°HCl(g) = -92,3 kJ/mol; ΔH°C2H2Cl4(l) = -333,36 kJ/mol

Replacing:

ΔH°r₂ = -276,2 kJ/mol -92,3 kJ/mol + 333,36 kJ/mol

<em>ΔH°r₂ = -35,14 kJ/mol</em>

<em></em>

I hope it helps!

4 0
3 years ago
Which one of the following solutions will have the greatest concentration of hydroxide ions?
jek_recluse [69]

Answer:

e = 0.250 M calcium hydroxide

it will produce 0.5 M [OH⁻].

Explanation:

A = 0.100 M HCl

HCl is an acid it would produce negligible amount of OH⁻.

B = 0.100 M magnesium hydroxide

Mg(OH)₂ + H₂O → Mg²⁺ + 2OH⁻

The ration of Mg(OH)₂  and OH⁻ is 1:2.

[OH⁻] = 2× 0.100 = 0.200 M

C =0.100 M ammonia

NH₃ + H₂O → NH₄OH → NH₄⁺ + OH⁻

The Kb expression will be written as,

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Kb = 1.8 ×10 ⁻⁵

Now we will put the values:

1.8 ×10 ⁻⁵ = [NH₄⁺] [OH⁻] / 0.100

[NH₄⁺] [OH⁻]  = x²

1.8 ×10 ⁻⁵ = x² / 0.100

x² = 1.8 ×10 ⁻⁵ × 0.100

x² =  1.8  × 10 ⁻⁶

x = 1.34 × 10 ⁻³

[OH⁻] = 1.34 × 10 ⁻³ M

d = 0.300 M rubidium hydroxide

RbOH + H₂O    →     Rb⁺ + OH⁻

The ratio of RbOH  and OH⁻ is 1:1.

[OH⁻] = 0.300 M

e = 0.250 M calcium hydroxide

Ca(OH)₂ + H₂O → Ca²⁺ + 2OH⁻

The ration of Ca(OH)₂  and OH⁻ is 1:2.

[OH⁻] = 2× 0.250 = 0.5 M

7 0
3 years ago
What is the pressure of a mixture of 0.200 g of H2, 1.00 g of N2 , and 0.820 g of Ar in a container with a volume of 2.00 L at 2
alexdok [17]

Answer:

P(mixture) = 1.92 atm

Explanation:

Given data:

Mass of H₂ = 0.200 g

Mass of N₂ = 1.00 g

Mass of Ar = 0.820 g

Volume = 2 L

Temperature = 20°C

Pressure of mixture = ?

Solution:

Pressure of hydrogen:

Number of moles of hydrogen = mass / molar mass

Number of moles of hydrogen = 0.200 g / 2 g/mol

Number of moles of hydrogen = 0.1 mol

P = nRT / V

P = 0.1 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 2.41 atm. L /2 L

P = 1.2 atm

Pressure of nitrogen:

Number of moles of nitrogen = mass / molar mass

Number of moles of nitrogen = 1 g / 28 g/mol

Number of moles of nitrogen = 0.04 mol

P = nRT / V

P = 0.04 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 0.96 atm. L /2 L

P = 0.48 atm

Pressure of argon:

Number of moles of argon = mass / molar mass

Number of moles of argon = 0.820 g / 40 g/mol

Number of moles of argon = 0.02 mol

P = nRT / V

P = 0.02 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 0.48 atm. L /2 L

P = 0.24 atm

Total pressure of mixture:

P(mixture)  = pressure of hydrogen + pressure of nitrogen + pressure of argon

P(mixture)  = 1.2 atm + 0.48 atm + 0.24 atm

P(mixture) = 1.92 atm

5 0
3 years ago
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