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3 years ago

How many molecules are there in 7.27 miles of ammonia?

Chemistry

2 answers:

worty [1.4K]3 years ago

5 0

Answer:

43.78 X 10^23

Explanation:

1 mole of NH3 has 6.023×10^23 molecules according to avogadro's principle

therefore, 7.27 moles = 6.023×10^23 X 7.27

= 43.78 X 10^23

NemiM [27]3 years ago

4 0

Hey There!

To solve this question remember that 1 mole of ammonia is equal to 17G of ammonia.

Using that, we first multiply 17 by 7 = 24

Then we multiply 7 by .27 = 1.89

Finally, taking those and adding them together gives you the answer...

25.89 grams :)

Have a great day, I hope I helped.

+ Brainliest & Five stars are always appreciated

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Answer:

8 Valence Electrons

Explanation:

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Sulfur has six valence electrons.
Therefore, there is a total of eight valence electrons in H2S.

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2 years ago

What is the concentration of an unknown Mg(OH)2 solution if it took an average of 15.4mL of

vova2212 [387]

Answer:

0.077M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HCl + Mg(OH)2 —> MgCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Concentration of base Cb =...?

Volume of base (Vb) = 10mL

Concentration of acid (Ca) = 0.1M

Volume of acid (Va) = 15.4mL

Step 3:

Determination of the concentration of the base, Mg(OH)2.

The concentration of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.1 x 15.4 /Cb x 10 = 2/1

Cross multiply to express in linear form

Cb x 10 x 2 = 0.1 x 15.4

Divide both side by 10 x 2

Cb = (0.1 x 15.4) /(10 x 2)

Cb = 0.077M

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7 0

3 years ago

A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad

Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -Weak acid-

1.0L × (0.050 mol / L) = 0.050 moles of NaF -Conjugate base-

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

(a) adding 0.050 mol of HCl: The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid destroying the buffer.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer wouldn't be destroyed.

3 0

3 years ago