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2 years ago

What is the potential energy of a rock that weights 100 Newtons that is sitting on top

Chemistry

1 answer:

weqwewe [10]2 years ago

6 0

Answer:

Us we know mass is constant any where .

Given :-

W=100N gravity=G=10 meter per second square

H=300M

W=M*G

w=m*10m/second square

100=M*10

M=10kg

P.E=M*G*H

P.E=10*10*300

P.E=30,000J

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Stells [14]

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2 years ago

True or False? The following set of quantum numbers is acceptable (0, 0, 0, +1/2). Explain.

leonid [27]

Answer:

False

Explanation:

The values of the principal quantum number range from 1- infinity. This implies that there is no such thing as an energy level with n=0.

Hence this can not be an acceptable set of quantum numbers when n=0.

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3 years ago

Carbon dioxide and water react to form bicarbonate ion and hydronium ion. Hyperventilation (rapid breathing) causes more carbon

Pani-rosa [81]

Answer:When a person hyperventilates they exhale more carbon dioxide than normal. As a result the carbon dioxide concentration in the blood is reduced and the bicarbonate/carbonic acid equilibrium shifts to the left. The corresponding drop in H3O+ concentration causes an increase in pH.

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2 years ago

Suppose 11.4g of ammonium chloride is dissolved in 250 ml of a 0.3M aqueous solution of potassium carbonate. Calculate the final

Sindrei [870]

Answer:

The molarity of the final ammonium cation is 0.252M

Explanation:

Step 1: Data given

Mass of ammonium chloride (NH4Cl) = 11.4 grams

Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L

Step 2: The balanced equation

2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3

Step 3: Calculate moles of (NH4)Cl

moles (NH4)Cl = 11.4 grams /53.49 g/mol

Moles (NH4)Cl = 0.213 moles

Step 4: Calculate moles of K2CO3

Moles K2CO3 = Molarity * Volume

Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles

Step 5: Calculate moles (NH4)Cl at the equilibrium

For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3

(NH4)2CO3l will dissolve in 2NH4+ + CO32-

Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles

Moles NH4+ = moles (NH4)Cl = 0.063 moles

Step 6: Calculate Molarity of NH4+

Molarity = Moles / volume

Molarity of NH4+ = 0.063 moles / 0.250 L

Molarity of NH4+ = 0.252 M

The molarity of the final ammonium cation is 0.252M

5 0

3 years ago

erik [133]

Answer:

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Explanation:

6 0

3 years ago