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2 years ago

During nuclear fission and fusion, what is matter that seems to disappear actually converted into?

Chemistry

2 answers:

viktelen [127]2 years ago

5 0

Answer:

The Correct Answer is " Energy"

Goodluck Guys

Explanation:

hjlf2 years ago

4 0

Answer:

During nuclear fission and fusion matter that seems to disappear but is actually converted into energy. The amount of energy (E) produced in such a reaction can be calculated using Einstein's formula for the equivalence of mass and energy: E = mc^2.

Explanation:

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Which is an example of gene modification that humans have done for thousands of years?

miv72 [106K]

Cross breeding plants to make better apples

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3 years ago

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Nadusha1986 [10]

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3 years ago

A reaction that is second-order in one reactant has a rate constant of 1 × 10–2 L/mol · s. If the initial concentration of the r

Harlamova29_29 [7]

Answer : It take time for the concentration to become 0.180 mol/L will be, 277.8 s

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 = $1\times 10^{-2}mol/L.s$

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[A] = concentration of substance after time 't' = 0.180 mol/L

$[A]_o$ = Initial concentration = 0.360 mol/L

Putting values in above equation, we get:

$1\times 10^{-2}=\frac{1}{t}\left (\frac{1}{0.180}-\frac{1}{0.360}\right)$

$t=277.8s$

Hence, it take time for the concentration to become 0.180 mol/L will be, 277.8 s

7 0

3 years ago

Ethanol has a Kb of 1.22 Degrees C/m and usually boils at 78.4 Degrees Celcius. How many mol of an nonionizing solute would need

Gala2k [10]

Answer:

0.3097 moles of an nonionizing solute would need to be added.

Explanation:

Molal elevation constant = $k_b=1.22^oC/m$

Normal boiling point of ethanol = $T_o=78.4^oC$

Boiling of solution = $T_b=86.30^oC$

Moles of nonionizing solute = n

Mass of ethanol (solvent) = 47.84 g

Elevation boiling point:

$\Delta T_b=T_b-T_o$

$\Delta T_b=86.30^oC-78.4^oC=7.9^oC$

$\Delta T_b=K_b\times m$

$m=\frac{\text{Moloes of solute}}{\text{Mass of solvent(kg)}}$

$7.9^oC=1.22^oC/m\times \frac{n}{0.04784 kg}$

n = 0.3097 mol

0.3097 moles of an nonionizing solute would need to be added.

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3 years ago

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Vladimir79 [104]

Answer:

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3 years ago

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