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kompoz [17]
2 years ago
6

If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?

Chemistry
1 answer:
ipn [44]2 years ago
5 0

Answer:

0.5667 M ≅ 0.57 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.

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100.mL of a .795 M solution of KBr is diluted to 500.mL. what is the new concentration of the solution?
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Answer:

0.159 M

Explanation:

convert from mL to L then use the equation:

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Why saturated fatty acid can’t create kink ?​
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A 6.40 g sample of a compound is burned to produce 8.37 g CO_2, 2.75 g H_2O, 1.06 g N_2, and 1.23 g SO_2. What is the empirical
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The empirical formula :

C₁₀H₁₆N₄SO₇

<h3>Further explanation</h3>

Given

6.4 g sample

Required

The empirical formula

Solution

mass C :

= 12/44 x 8.37 g

= 2.28

mass H :

= 2/18 x 2.75 g

= 0.305

mass N = 1.06

mass S :

= 32/64 x 1.23

= 0.615

mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g

Mol ratio :

= C : H : N : S : O

= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16

= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019

= 10 : 16 : 4 : 1 : 7

The empirical formula :

C₁₀H₁₆N₄SO₇

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