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Maru [420]
2 years ago
11

If a radioactive decay produced 1.26 × 1012 J of energy, how much mass was lost in the reaction?

Chemistry
1 answer:
Ket [755]2 years ago
8 0

The mass lost in the process is 1.4 * 10^-5Kg.

<h3>What is the Einstein equation?</h3>

The Einstein equation is able to relate the mass lost to the energy that is evolved in a nuclear process.

Given that;

E = Δmc^2

Δm = E/c^2

Δm =  1.26 × 10^12 J/( 3 * 10^8)^2

Δm = 1.4 * 10^-5Kg

Learn more about Einstein equation:brainly.com/question/10809666

#SPJ1

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Calculate the pHpH of a 0.10 MM solution of HClHCl . Express your answer numerically using two decimal places.
alisha [4.7K]

Answer: The pH 0f 0.10 solution of HCl is 1.00

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

HCl\rightarrow H^++Cl^{-}

According to stoichiometry,  

1 mole of HCl gives 1 mole of H^+  

Thus 0.10 moles of HCl gives =\frac{1}{1}\times 0.10=0.10 moles of H^+  

Putting in the values:

pH=-\log[0.10]

pH=1.00

Thus pH 0f 0.10 solution of HCl is 1.00

8 0
3 years ago
__HgO __Hg + __O2 what is the reaction
Gelneren [198K]

Answer:

2HgO <---- 2Hg + O2

Explanation:

8 0
3 years ago
Define a transverse wave
Mumz [18]
Transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
4 0
3 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
Use the solubility generalizations on the information page to predict if one or more precipitates will form when aqueous solutio
neonofarm [45]

Answer:

3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)

Explanation:

In solubility rules, all ammonium and nitrates ions are solubles and all sulfates are soluble except the sulfates that are produced with Ca²⁺, Sr²⁺, Ba²⁺, Ag⁺ and Pb²⁺. That means the NH4NO3 and the Al2(SO4)3 produced are both <em>soluble and no precipitate is predicted. </em>

The reaction is:

<h3>3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)</h3>
6 0
3 years ago
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