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Serhud [2]
3 years ago
14

If E=mc^2,solve for both m and c. Also, if m=80 and c=0.40 what is the value of E?

Chemistry
1 answer:
Pie3 years ago
4 0

<u><em>Answer:</em></u>

m = \frac{E}{c^2}

c = \sqrt{\frac{E}{m} }

E = 12.8 J

<u><em>Explanation:</em></u>

<u>Part 1: Solving for m</u>

<u>We are given that:</u>

E = mc²

To solve for m, we will need to isolate the m on one side of the equation

This means that we will simply divide both sides by c²

m = \frac{E}{c^2}

<u>Part 2: Solving for c</u>

<u>We are given that:</u>

E = mc²

To solve for c, we will need to isolate the m on one side of the equation

This means that first we will divide both sides by m and then take square root for both sides to get the value of c

c^2 = \frac{E}{m}\\  \\ c=\sqrt{\frac{E}{m}}

<u>Part 3: Solving for E</u>

<u>We are given that:</u>

m = 80 and c = 0.4

<u>To get the value of E, we will simply substitute in the given equation: </u>

E = mc²

E = (80) × (0.4)²

E = 12.8 J

Hope this helps :)

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irga5000 [103]

Sodium carbonate is used to neutralised sulfuric acid, H₂SO₄. Sodium carbonate is the salt of stron base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

Na₂CO₃ + H₂SO₄→ Na₂SO₄ + H₂CO₃

From balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄. Molar mass of Na₂CO₃= 106 g/mol=0.106 kg/mol and molar mass of H₂SO₄= 98 g/mol=0.098 kg/mol.

To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, to neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is=\frac{0.106 X 5.04 X 10^{3} }{0.098} kg= 5.451 X 10³ kg.

5 0
3 years ago
An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .
grandymaker [24]

Answer: The specific heat of the unknown metal is 0.897J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=5040 Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = T_i

Final temperature of the water = T_f

Change in temperature ,\Delta T=T_f-T_i=(64.7)^0C

Putting in the values, we get:

5040=86.8\times c\times 64.7^0C

c=0.897J/g^0C

The specific heat of the unknown metal is 0.897J/g^0C

4 0
3 years ago
A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa
Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

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Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

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