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Advocard [28]
2 years ago
3

Johnny wished to create a 0.300 M NaCl solution. Johnny only had 50.0 mL of water, however. To create his goal solution, Johnny

knew he needed to know the molar mass of NaCl, which was . Next, to find the the number of moles of NaCl he needed, Johnny simply multiplied the by the volume of water, in liters, which was L. Then, he calculated the number of grams of NaCl he needed taking this number of moles of NaCl and the molar mass. Finally, he weighed out this number of grams of NaCl and dissolved the salt in to create the 0.300 M solution.
Chemistry
1 answer:
kykrilka [37]2 years ago
7 0

Answer:

1. Mole of NaCl = 0.015 mole.

2. Molar Mass of NaCl = 58.5g/mol

3. Mass of NaCl = 0.88g

Explanation:

Step 1:

Data obtained from the question.

Molarity of NaCl = 0.3M

Volume = 50mL = 50/1000 = 0.05L

Number of mole =?

Molar Mass of NaCl =?

Mass of NaCl =.?

Step 2:

Determination of the number of mole of NaCl. This is illustrated below:

Molarity = mole of solute /Volume of solution

Mole of solute = molarity x volume

Mole of NaCl = 0.3 x 0.05

Mole of NaCl = 0.015 mole.

Step 3:

Determination of the molar mass of NaCl.

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Step 4:

Determination of the mass of NaCl.

Mole of NaCl = 0.015 mole

Molar Mass of NaCl = 58.5g/mol

Mass of NaCl =?

Mass = number of mole x molar Mass

Mass of NaCl = 0.015 x 58.5

Mass of NaCl = 0.8775g ≈ 0.88g

Therefore, Johnny weighed 0.88g of NaCl and dissolved it in 50mL of water to produce 0.3M NaCl solution

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Luba_88 [7]
I don't know if you didn't gave a picture choice or if i didn't get the picture.
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7 0
2 years ago
Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod
son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

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2 years ago
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sertanlavr [38]

Answer:

Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.

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2 years ago
Given that the density of TlCl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, determine how many form
ryzh [129]
Here's my best guess
the volume of the unit cell is (385*10^-12)^3=5.7066*10^-29 m^3
multiply by density to get mass mass = (7 g/cm^3)*(100^3 cm^3 / 1^3 m^3) * 5.7066*10^-29 m^3= 3.99466*10^-22 g
covert to moles 3.99466*10^-22 g * 1 mol / 239.82 g = 1.6657 *10^-24 mol
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1.6657 *10^-24 mol * 6.23*10^23 units/mol = 1.04

385 pm = 3.85*10^(-8) cm

The volume of the unit cell is the cube of that, which is 5.71*10^(-23) cm^3. Since the ratio of mass to volume (i.e. the density) must be the same no matter what amount of TlCl you have, you can say:
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Explanation:

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