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3 years ago

How many joules of energy are made by the loss of 100kg of mass using the formula E=mc2? please help!!!

Chemistry

1 answer:

asambeis [7]3 years ago

6 0

Answer:

E = 3 × 10¹⁰ J

Explanation:

Mass, m = 100 kg

We need to find energy made by the loss of 100 kg of mass. The formula between the mass and energy is given by :

E = mc²

Where c is speed of light

Putting all the values, we get :

E = 100 kg × (3×10⁸ m/s)²

= 3 × 10¹⁰ J

So, the required energy is 3 × 10¹⁰ J.

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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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3 years ago

Which excerpt from “Seventh Grade” is an example of an internal expectation?

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3 years ago

How many kL does a 9.51 ´ 109 cL sample contain?

elena-14-01-66 [18.8K]

Answer:

9.51 × 10⁴ kL

Explanation:

Step 1: Given data

Volume of the sample (V): 9.51 × 10⁹ cL

Step 2: Convert "V" to liters

We will use the conversion factor 1 L = 100 cL.

9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L

Step 3: Convert "V" to kL

We will use the conversion factor 1 kL = 1000 L.

9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL

9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.

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