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Jobisdone [24]
3 years ago
15

A nuclear fisston reaction has mass difference between the products and the reactants of 3,86 . Calculate the amount of energy r

eleased by the reaction . A) 1.16 * 10 ^ 6 * J B) 2.59 * 10 ^ 9 3.47 * 10 ^ 14 * j D 4.82*10^ 17 )
Chemistry
1 answer:
Savatey [412]3 years ago
8 0

The question is incomplete, here is the complete question:

A nuclear fisston reaction has mass difference between the products and the reactants of 3.86 g. Calculate the amount of energy released by the reaction.

A) 1.16 * 10^6 * J

B) 2.59 * 10^9

C) 3.47 * 10^ 14 * J

D) 4.82 * 10^17 J

<u>Answer:</u> The amount of energy released by the reaction is 3.47\times 10^{14}J

<u>Explanation:</u>

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

where,

\Delta m = mass defect = 3.86 g = 0.00386 kg    (Conversion factor: 1 kg = 1000 g)

c = speed of light = 3\times 10^8m/s

Putting values in above equation, we get:

E=(0.00386kg)\times (3\times 10^8m/s)^2

E=3.47\times 10^{14}J

Hence, the amount of energy released by the reaction is 3.47\times 10^{14}J

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Answer:

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Explanation:

Given that,

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When of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing poi
Nesterboy [21]

The given question is incomplete. The complete question is:

When 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of ammonium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X.

Answer:  the van't Hoff factor for ammonium chloride is 1.74

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point  

K_f = freezing point constant = ?

i = 1 ( for non electrolyte)

m= molality

8.2^0C=1\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (glycine) = 75.07 g/mol

Mass of solute (glycine) = 282 g

8.2^0C=1\times K_f\times \frac{282g}{75.07g/mol\times 0.95kg}

K_f=2.07

ii) 20.0^0C=i\times \times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (ammonium chloride) = 53.49 g/mol

Mass of solute (ammonium chloride) = 282 g

20.0^0C=i\times 2.07\times \frac{282g}{53.49g/mol\times 0.95kg}

i=1.74

Thus the van't Hoff factor for ammonium chloride is 1.74

4 0
3 years ago
How do I go from molecules to moles? For example 1.50e23 molecules NH3 how many moles are in this problem?
maxonik [38]
To go from molecules to moles divide by Avogadro's number [<span>6.02x10^23]</span>
Example:
1.50x10^23 divided by 6.02x10^23 = 0.249 moles rounded to three significant figures
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3 years ago
Someone please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
OLga [1]

Answer:

1 - B

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