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gavmur [86]
3 years ago
10

The radioactive atom R 88 210 a is an alpha emitter. What nucleus does it produce?

Chemistry
1 answer:
olasank [31]3 years ago
5 0

Answer:

X 86 206

Explanation:

Radioactive atoms are nuclei that can under go disintegration to emit either an alpha particle, beta particle or gamma radiation. The process could be spontaneous or stimulated.

When a radioactive atom R 88 210 emits alpha particle, it would produce an element with atomic number 86 and mass number 206 i.e X 86 206. An alpha particle is usually a helium nucleus.

                     R^{210} ⇒ x^{206} + He^{4} + energy

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bekas [8.4K]

Explanation:

2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant of reaction = K=154

Concentration of NO = [NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M

Concentration of bromine gas = [Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M

Concentration of NOBr gas = [Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M

The reaction quotient is given as:

Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}

Q=328.06

Q>K

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced.  False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0
3 years ago
How many joules of heat must be transferred to a 480-g aluminum pizza pan to raise its temperature from 22C to 234C? The specifi
NemiM [27]
The given formula for heat, Q=mc(Tf-Ti), is the best way to solve such problems with changes in temperature. It can be said that m is the mass of the substance. C is the specific heat of the substance. The term (Tf-Ti) is the change in temperature.

Q = mc(Tf-Ti) = 480g(0.96 J/g-C)(234-22) = 97689.6 Joules of heat
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Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0
3 years ago
The acid dissociation constant ka for an unknown acid ha is 4.57 x 10^-3 what is the base dissociation constant kb for th econju
SashulF [63]

Answer:

2.19 x 10^-12.

Explanation:-

The relation between Ka and Kb for an acid and it's conjugate base is

Ka x Kb = Kw where Kw = ionic product of water.

So Kb = 10^-14 / (4.57 x 10 ^ -3)

= 2.19 x 10^-12

4 0
3 years ago
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