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vfiekz [6]
2 years ago
8

How many joules of energy are produced when 3.0 × 10-28 kilograms of mass are lost? [(1 j = 1 ) and the speed of light = 3.00 ×

108 ]
Chemistry
1 answer:
Aleksandr [31]2 years ago
8 0

The equation that we are going to use on this problem is the famous Einstein field<span> equation. E = mc^2 where E is the energy (to be computed), m is the mass (</span>3.0 × 10-28 kilograms) and c is the speed of light (3.00 × 10^8). If we plug in the given into the equation the answer will be 2.7*10^-11 KJ or <span>2.7*10^-8 J. </span>

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How many particles are present in0.24moles of carbon?
Ray Of Light [21]

Answer:

1.45 x 10²³ particles

Explanation:

Given parameters:

Number of moles of carbon  = 0.24moles

Unknown:

Number of particles = ?

Solution:

A mole of a substance contains the Avogadro's number of particles.

 The Avogadro's number of particles is 6.02 x 10²³

So;

  0.24 moles of carbon will contain 0.24 x 6.02 x 10²³  = 1.45 x 10²³ particles

7 0
2 years ago
What is an alkaline earth metal that has 5 shells?
brilliants [131]

Answer:

The elements in the alkaline earth metals group; beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra), have two electrons in their outer electronic shell.

Explanation:

3 0
3 years ago
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The reaction H2SO4 + 2 NaOH -&gt; Na2SO4 + 2H2O represents an acid-base tritration. The equivalence point occurs when 24.75 mL o
artcher [175]
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Hence
            [H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
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6 0
3 years ago
In which one of the following branches of natural science are properties of materials studied?
Leya [2.2K]

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Chemistry maybe

Explanation:

3 0
3 years ago
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The osmotic pressure, π, of a solution of glucose is 132 atm . find the molarity of the solution at 298 k.
Marina86 [1]

The formula for osmotic pressure is:

\Pi = iMRT

where \Pi is osmotic pressure, i is van't Hoff's factor, M molarity, R is Ideal gas constant, and T is Temperature.

\Pi = 132 atm

The van't Hoff's factor for glucose, i = 1

R = 0.08206 Latmmol^{-1}K^{-1}

T = 298 K

Substituting the values in the above equation we get,

132 atm = 1\times M\times 0.08206 Latmmol^{-1}K^{-1}\times 298

M = \frac{132 atm}{1\times 0.08206 Latmmol^{-1}K^{-1}\times 298} = 5.4797 molL^{-1} \simeq 5.48 molL^{-1}

So, the molarity of the solution is 5.48 molL^{-1}.

4 0
3 years ago
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