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Elden [556K]
3 years ago
8

What is the molarity of a solution prepared by dissolving 30.0g NaCl in enough water to make a 0.300 L solution?

Chemistry
1 answer:
meriva3 years ago
5 0

Answer:

D.) 1.71 M NaCl

Explanation:

Molarity equation: M= n/v

n= moles of solute

v=liters of solution

NaCl= 58.443 g/mol

30g NaCl / 58.443g/mol = 0.5133(this is n)

0.5133 mols/0.300 L=1.71115674 M

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Answer:

B = A/DH – C

Explanation:

From the question given above, we obtained:

A = D • H(B + C)

Thus, we can obtain B in terms of D, H, A and C by doing the following:

A = D • H(B + C)

A = DH(B + C)

Divide both side by DH

A/DH = B + C

Subtract C from both side

A/DH – C = B + C – C

A/DH – C = B

B = A/DH – C

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ANSWER IS (A)

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Bronsted-Lowry concept states that a substance is an acid if it can act as a H+ donor.

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However, Methanol (CH3OH) its written that it is liquid, i.e. pure methanol, CH3OH(l).  It is both acidic as well as basic. when it is mixed with water then it behaves as an acid.

The last one ammonia in gas phase is also neutral because its not in water. if mixed in water it behaves as a base.



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