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3 years ago

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The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a

mishap, what mass of the isotope will remain after 49.2 years?

1 answer:

Amiraneli [1.4K]3 years ago

3 0

Answer:

Explanation:

it is tough question plz give me some time i would give you your ans soon

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How does molecular attraction affect the amount of energy needed to create a phase change. Think about low, medium, and high

Art [367]

Answer:

When considering phase changes, the closer molecules are to one another, the stronger the intermolecular forces. Good! For any given substance, intermolecular forces will be greatest in the solid state and weakest in the gas state.

In the case of melting, added energy is used to break the bonds between the molecules. ... If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.

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3 years ago

How many significant figures in 11 soccer players

maria [59]

Counting gives an exact number and exact numbers have infinite sig figs.

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4 years ago

Structure can affect the Ka values for related acids. In the boxes below, draw the complete Lewis structure for a single ion of

Zigmanuir [339]

Answer:

Explanation: see attachment below

8 0

3 years ago

The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati

a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.80M^{-1}s^{-1}$

t = time taken = 95 s

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

$0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)$

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0

3 years ago

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

adell [148]

Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

$CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ$ ...[1]

Step 2 :

$6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ$ ..[2]

Adding 6 × [1] and [2]:

$6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)$

$6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

we get :

$6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?$

$\Delta H'=6\times \Delta H_1+\Delta H_2$

$\Delta H'=6\times 414.0 kJ+132.0kJ$

$\Delta H'=2,626 kJ$

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

$\frac{ 2,626 kJ}{5 } = 523.2 kJ$

7 0

3 years ago