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Sever21 [200]
3 years ago
6

What are the products in the equation for cellular respiration? A. Oxygen and lactic acid b. Carbon dioxide and water c. Glucose

and oxygen d. Water and glucose
Chemistry
1 answer:
Colt1911 [192]3 years ago
4 0
<span>Cellular respiration is the chemical reaction in which glucose and oxygen are turned into water, carbon dioxide, and energy. Now you know the answer. :P</span>
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The molecular weight is the same as molecular mass. You find this by using the periodic table 
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3 years ago
Help me, I’m failing chem
ladessa [460]

Answer:

b

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5 0
3 years ago
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Match each element to the number of electrons in its valence shell.
irinina [24]

Answer:

C: 4

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1 year ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

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3 years ago
Which of these is a property of bases?
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They turn litmus paper blue
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