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Setler79 [48]
3 years ago
7

A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th

e solution if the density is 1.12 g/mL.
Physics
1 answer:
Lyrx [107]3 years ago
5 0

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass =volume\times density

Using the formula

Mass of solution=1.12\times 473=529.76 g

Mass % of KBr=\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100

Mass % of KBr=\frac{49.3}{529.76}\times 100

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

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1,000,000
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Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1
My name is Ann [436]

The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons}  = 4.41294 Mev/nucleon

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3 0
3 years ago
How Many Negative (-) Electrons are there
makvit [3.9K]

Answer:

All electrons are negative(-) charged

5 0
3 years ago
Frank wrote several statements to summarize the relationship between the first and second laws of thermodynamics.
Kaylis [27]
First law is Conservation of Energy

Second is that entropy of an isolated system will always increase with time.

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4 years ago
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A 1.0 C charged object and a 2.0 C charged object are separated by 100 m. Where should a -1.0x10-3 C charged object be placed on
horsena [70]

Answer:

x = 41.2 m

Explanation:

The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.

In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract

             F = F₁₃ - F₂₃

let's write the expression for each force, let's set a reference frame on the charge q1

           F₁₃ = k \frac{q_1 q_3}{x^2}

           F₂₃ = k \frac{q_2 q_3}{(d-x)^2}

they ask us that the net force be zero

           F = 0

           0 = F₁₃ - F₂₃

           F₁₃ = F₂₃

          k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}

          \frac{q_1}{x^2} = \frac{q_2}{(d-x)^2 }q1 / x2 = q2 / (d-x) 2

       

           (d-x)² = \frac{q_2}{q_1} x²

we substitute

           (100 - x)² = 2/1  x²

           100- x = √2  x

           100 = 2.41 x

           x = 41.2 m

6 0
3 years ago
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