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3 years ago

11

Write the equation for the decay of kr-85

2 answers:

iris [78.8K]3 years ago

8 0

Answer:see image attached

Explanation:

Kr-85 turns spontaneously into a Rubidium nucleus by emitting a beta particle and a gamma ray. The rubidium nucleus has the same mass as the krypton nucleus. The gamma rays carries off the excess energy released. It is actually a photon of light. A beta particle is actually an electron. A lot of energy is usually released in this process.

Ray Of Light [21]3 years ago

7 0

<u>Answer</u>:-

Kr-85 changes into Rb-86 by mostly beta decay.

Equation:

85Kr → Rb-86 + β-particle + Energy

Explanation:

1)Link:

http://www.nucleide.org/DDEP_WG/Nuclides/Kr-85_tables.pdf

2) Photo

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How much heat is gained when a 50.32g piece of aluminum is heated from 9.0°c to 16°c

Rashid [163]

Answer: 317 joules

Explanation:

The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

In this case,

Q = ?

Mass of aluminium = 50.32g

C = 0.90J/g°C

Φ = (Final temperature - Initial temperature)

= 16°C - 9°C = 7°C

Then, Q = MCΦ

Q = 50.32g x 0.90J/g°C x 7°C

Q = 317 joules

Thus, 317 joules of heat is gained.

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3 years ago

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What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

rewona [7]

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$

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Answer:

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Explanation:

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3 years ago

A piece of magnesium is in the shape of a cylinder with a height of 6.62 cm and a diameter of 2.34 cm. If the magnesium sample h

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Answer:

0.600 g/cm³

Explanation:

Step 1: Given data

Height of the cylinder (h): 6.62 cm

Diameter of the cylinder (d): 2.34 cm

Mass of the cylinder (m): 17.1 g

Step 2: Calculate the volume of the cylinder

First, we have to determine the radius, which is half of the diameter.

r = d/2 = 2.34 cm/2 = 1.17 cm

Then, we use the formula for the volume of the cylinder.

V= π × r² × h

V= π × (1.17 cm)² × 6.62 cm

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Step 3: Calculate the density (ρ) of the sample

The density is equal to the mass divided by the volume.

ρ = m/V

ρ = 17.1 g/28.5 cm³

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3 years ago

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3 years ago

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