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Volgvan
2 years ago
14

How many molecules are in 1.00 g of C10H14N2? 3.71 x 1021 9.76 x 1025 1.02 x 10-26 2.32 x 1022

Chemistry
1 answer:
Reika [66]2 years ago
8 0

Answer:

3.71 x 1021 (First option)

Explanation:

Let's determine the molar mass of the compound.

C₁₀H₁₄N₂ = 12. 10 + 14 .1 + 14 .2 = 162 g/mol

So 162 grams of compound are contained in 1 mol

1 g of  C₁₀H₁₄N₂ are contained in ( 1 / 162) = 0.00617 moles

Let's conver the moles into amount of molecules (mol . NA)

0.00617 . 6.02×10²³ = 3.71×10²¹ molecules

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Read 2 more answers
A sample of gas has a mass of 0.560 g . Its volume is 125 mL at a temperature of 85 ∘C and a pressure of 757 mmHg .
-BARSIC- [3]

Answer:

132g/mole

Explanation:

using the formula PV=nRT should be used to solve for the number of moles (n).  R is a constant which is 62.3637 L mmHG/mole K.

Inorder for your units to match you will have to convert 125ml to .125L and the temperature of 85C to K . you do that by adding 273 to the 85C and get 358K.  Once you solve for n then you use that number and divide by the number of grams from the question (.560g) since molar mass is grams/moles.

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3 years ago
Help me with this science question on the middle one..
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8 0
2 years ago
n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form
Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

pK_a=8.0

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}

Now put all the given values in this expression, we get:

6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}

\frac{[Deprotonated]}{[Protonated]}=0.01  

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

\frac{[Protonated]}{[Deprotonated]}=100  

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0
2 years ago
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