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3 years ago

Write the equation for the decay of kr-85

Chemistry

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

Answer:-

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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an unknown molecule is found to consist of 24.2% carbon by mass, 4.0% hydrogen by mass and the remaining mass is due to chlorine

Paha777 [63]

Answer:

C3 H6 Cl 3

Explanation:

C -24.2%

H - 4.0%

Cl - (100-24.2 - 4.0)=73.8 %

We can take 100g of the substance, then we have

C -24.2 g

H - 4.0 g

Cl - 73.8 g

Find the moles of these elements

C -24.2 g/12.0 g/mol =2.0 mol

H - 4.0 g/1.0 g/mol = 4. 0 mol

Cl - 73.8 g/ 35.5 g/mol = 2.1 mol

Ratio of these elements gives simplest formula of the substance

C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1

CH2Cl

Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol

Real molar mass = 150 g/mol

real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3

So, Real formula should be C3 H6 Cl 3.

4 0

3 years ago

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Kazeer [188]

Answer:

what is the formula of rectangle

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2 years ago

The density of water is 1 g/cm3(1 gram per cubic centimeter); using this fact as a reference & Table c , how would you deter

NISA [10]

Answer:

Explanation:

Any densities less than 1g/cm3 will float, while objects with densities over 1g/cm3 will sink.

6 0

2 years ago

An exhaled air bubble underwater at 290.

Vinil7 [7]

The answer for the following problem is mentioned below.

Therefore the final volume of the gas is 52.7 ml.

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 290 kPa

Final pressure ( $P_{2}$ ) = 104 kPa

Initial volume ( $V_{1}$ ) = 18.9 ml

To find:

Final volume ( $V_{2}$ )

We know;

From the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

P × V = constant

P ∝ $\frac{1}{V}$

From the above equation;

$\frac{P_{1} }{P_{2} } = \frac{V_{2} }{V_{1} }$

$P_{1}$ represents the initial pressure of the gas

$P_{2}$ represents the final pressure of the gas

$V_{1}$ represents the initial volume of the gas

$V_{2}$ represents the final volume of the gas

Substituting the values of the above equation;

$\frac{290}{104}$ = $\frac{V_{2} }{18.9}$

$V_{2}$ = 52.7 ml

Therefore the final volume of the gas is 52.7 ml.

6 0

2 years ago

If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised

alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}$

Solving for V₂ , we get:

V₂ = 15.04 mL

3 0

3 years ago