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stepan [7]
3 years ago
9

If there is a loss of 5.63 x 10-7kg of mass in a nuclear reaction, how many Joules of energy would be released? Recall that c =

3x108 m/s.
Chemistry
1 answer:
Fudgin [204]3 years ago
5 0
Thank you for posting your question here at brainly. 

E = mc^2 
<span>where E is the energy in joules, </span>
<span>m is the mass in kilograms, </span>
<span>and c is the speed of light. </span>

<span>E = mc^2 </span>
<span>E = (5.63 x 10^-7 kg)(3 x 10^8 m/s)^2 </span>
<span>E = 5.07 x 10^10 J </span>


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What is the strongest interparticle force in a sample of kr?
Anit [1.1K]
Answer:
            London Dispersion Force<span> is the strongest interparticle force in a sample of Kr.

Explanation:
                   Krypton belongs to Noble Gases. They exists in monoatomic form as they are inert in nature and are very less reactive. So, as there is no polarity in Krypton so it will fail to create either Dipole-Dipole or Hydrogen Bond Interactions between its atoms. While, London Dispersion Forces or Van Der Waals forces can exist between Kr atoms. When Kr atoms approaches one other they create Instantaneous dipole. This</span> Instantaneous dipole induces dipole in second Kr atom and the process starts propagating. Hence, interactions are generated between Kr atoms.
7 0
3 years ago
You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of
Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}

Amount of BaSO₄ from K₂SO₄ is;

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4} = X\times\frac{233.38 }{142.04} = 1.643·X

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4} = Y\times\frac{233.38 }{174.259 } = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0
3 years ago
A 7.50-g sample of iron is heated in oxygen to form an iron oxide. if 10.71 g of the oxide is formed, what is its empirical form
Zielflug [23.3K]
Answer is: empirical formula is Fe₂O₃.
m(Fe) = 7,50 g.
m(iron oxide) = 10,71 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 7,50 g ÷ 55,85 g/mol = 0,134 mol.
m(O) = m(iron oxide) - m(Fe).
m(O) = 10,71 g - 7,50 g = 3,21 g.
n(O) = 3,21 g ÷ 16 g/mol = 0,20 mol.
n(Fe) : n(O) = 0,134 mol : 0,2 mol = 2 : 3.
3 0
3 years ago
000333 g fluorescein (332.32 g/mol) is dissolved in 225 ml solution of ethanol. the density of ethanol is 0.785 g/ml. what is th
andrey2020 [161]
1) Molarity

M = n / V
n: number of moles of solute
V: volume of the solution in liters

n = mass / molar mass = 0.000333 g / 332.32 g / mol =  1*10 ^ - 6 moles

V = 225 ml * 1 liter / 1000 ml = 0.225 liter

M = 10^-6 mol / 0.225 liter = 0.00000444 M

2) ppm

ppm = parts per million

grams of solute: 0.000333 g

grams of solution = volume * density = 225 ml * 0.785 g / ml = 176.625 g

ppm = [0.00033 g / 176.625 g] * 1,000,000 = 1.868 ppm

 

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2 years ago
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They all face the characteristics of life, they are all living, and they all have cells.
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3 years ago
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