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3 years ago

16.0 grams of molecular oxygen gas is equal to

Chemistry

2 answers:

julsineya [31]3 years ago

7 0

Answer:A

Explanation:

IceJOKER [234]3 years ago

4 0

One thing to notice in the question is, we are asked about molecular oxygen that has formula O2 not atomic oxygen O.

As we are asked about molecular oxygen, we will answer the question in terms of number of molecules that are present in 16 grams of molecular oxygen.

To get the number of molecules present in 16 grams of O2, we will use the formula:

No. of molecules = no. of moles x Avogadro's number (NA)----- eq 1)

As we know:

The number of moles = mass/ molar mass of molecule

Here we have been given mass already, 16 grams and the molar mass of O2 is 32 grams.

Putting the values in above formula:

= 16/32

= 0.5 moles

Putting the number of moles and Avogadro's number (6.02 * 10^23) in eq 1

No. of molecules = 0.5 x 6.02 * 10^23

=3.01 x 10^23 molecules

or 301,000,000,000,000,000,000,000 molecules

This means that 16 grams of 3.01 x 10^23 molecules of oxygen.

Hope it helps!

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Artemon [7]

Answer:

$9.8ms^{-2}$

Explanation:

using the equation :

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where v is 98m/s

u is 0m/s

t is 10seconds

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3 years ago

Wetlands provide habitats for many living things because of their sheltered waters and

777dan777 [17]

The choices can be found elsewhere and as follows:

a. large supply of nutrients.
b. thick layer of mud.
c. lack of insects
d. ability to prevent floods.

I believe the correct answer is option A. Wetlands provide habitats for many living things because of their sheltered waters and large supply of nutrients. Hope this answers the question.

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3 years ago

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Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

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3 years ago

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Papessa [141]

Explanation:

SADMEP

-2(bx-5) = 16 distribute

-2bx +10 = 16 subtracte

-10 -10

-2bx = 6

divide by -2x (on both sides)

b = -3x

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3 years ago

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Fudgin [204]

Answer:

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Explanation:

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