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3 years ago

How many moles of nacl are required to prepare 0.80 l of 6.4 m nacl? 0.13 mol nacl 5.1 mol nacl 7.2 mol nacl 8.0 mol nacl?

Chemistry

2 answers:

fomenos3 years ago

5 0

NaCl = n/V

6.4M = n/0.80L 

n= 5.12mol

dangina [55]3 years ago

3 0

Answer:

5.12 moles of NaCl are required to prepare 0.80 L of 6.4 M NaCl.

Explanation:

Molarity (M) is a concentration measure that indicates the number of moles of solute that are dissolved in a given volume.

The Molarity is then determined by:

$Molarity (M)=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

6.4 M NaCl indicates that 1 liter of solution there are 6.4 moles of NaCl.

You can apply a rule of three as follows: if in 1 liter of solution there are 6.4 moles of NaCl, in 0.8 L how many moles are there?

$moles=\frac{0.8 L*6.4 moles}{1 L}$

moles=5.12

5.12 moles of NaCl are required to prepare 0.80 L of 6.4 M NaCl.

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2 years ago

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In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

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The given electronic configuration is: $1s^22s^22p^63s^2$

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The given electronic configuration is: $1s^22s^22p^63s^3$

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The number of valence electrons in the given configuration are [2 + 6] = 8

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Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

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