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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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Seawater is typically 3.5% salt and has a density of 1.03 g/mL. How many grams of salt would be needed to prepare enough seawate

Bas_tet [7]

Answer:

Amount of salt needed is around 2.3*10³ g

Explanation:

The salt content in sea water = 3.5 %

This implies that there is 3.5 g salt in 100 g sea water

Density of seawater = 1.03 g/ml

Volume of seawater = volume of tank = 62.5 L = 62500 ml

Therefore, the amount of seawater required is:

$=Density*Volume = 1.03g/ml*62500ml = 6.44*10^{4} g$

The amount of salt needed for the calculated amount of seawater is:

$=\frac{6.44*10^{4}g\ water*3.5g\ salt }{100g\ water} =2254 g =2.3*10^{3} g$

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3 years ago

Which is one way that topographic maps are different from other maps?

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3 years ago

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Why does the water turn blue after some time when blue ink is dropped in water? Which property

zepelin [54]

Answer:

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Explanation:

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3 years ago

What is the molar mass of an element?

Vesnalui [34]

The answer is: the mass of 6.02 x 1023 representative particles of the element.

The base SI unit for molar mass is kg/mol, but chemist more use g/mol (gram per mole).

For example, molar mas of ammonia is 17.031 g/mol.

M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.

M(NH₃) = 14.007 + 3 · 1.008 · g/mol.

M(NH₃) = 17.031 g/mol.

The molar mass (M) is the mass of a given substance (in this example ammonia) divided by the amount of substance.

8 0

3 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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3 years ago