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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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Answer:

3.01×10²³ atoms of calcium

Explanation:

number of moles = number of atoms/Avogadro's constant

n = N/NA

N = n×NA = 0.500 mol×6.02×10²³ mol^-1

N = 3.01×10²³ atoms of calcium

3 0
3 years ago
The limiting reactant, O2, can form up to 2.7 mol Al2O3. What mass of Al2O3 forms?
Margaret [11]

Answer:

280 g Al₂O₃

Explanation:

To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).

Molar Mass (Al₂O₃): 101.96 g/mol

2.7 moles Al₂O₃          101.96 g
------------------------  x  -------------------  = 275 g Al₂O₃  = 280 g Al₂O₃
                                     1 mole

5 0
2 years ago
3. Briefly discuss the results of the TLC. Was there evidence of unreacted p-nitrobenzaldehyde in either product
Tems11 [23]

TLC means Thin Layer Chromatography. It is a method that can best be described as "Affinity-Based" used in the separation of compounds that are in a mixture.

<h3>What is unreacted p-nitrobenzaldehyde?</h3>

Unreacted p-nitrobenzaldehyde is simply an organic aromatic compound that contains a nitro group para-substituted to an aldehyde. in this case, if it is unreacted, that means it is the same as before the chemical reation.

Note that the question is missing key information hence the general answer.

Learn more about TCL at:

brainly.com/question/10296715

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2 years ago
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Explanation :

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The expression for rate of reaction for the reactant :

\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}

\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}

The expression for rate of reaction for the product :

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\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}

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3 0
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Answer:

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