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ololo11 [35]
3 years ago
8

Of the concentration units below, only ________ uses kg of solvent in its calculation.

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
6 0

Answer:

E) molality

Explanation:

Molality -

Molarity of a substance , is the number of moles present in a Kg of solvent .

Hence , the formula for molality is given as follow -

m = n / s

m = molality

s = mass of solvent in Kg ,

n = moles of solute ,

Hence , from the given information of the question,

The concentration unit which have Kg of solvent , is molality.

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Heat Transfer Lab

The following represents a lab set up for heat transfer. The cup on the left started with boiling water at 100 degrees C and the cup on the right has water at 20 degrees C. There is an aluminum bar between the two cups allowing heat to transfer from one cup into the other. The set up will be left alone for 20 minutes and temperatures of each cup of water will be recorded every minute for 20 minutes.

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3 years ago
How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?
algol [13]

Answer:

1.445\times 10^{23} formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = \frac{13.0 g}{162.5 g/mol}=0.08 mol

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

3\times 0.08 mol=0.24 mol of chloride

Na^++Cl^-\rightarrow NaCl

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

\frac{1}{1}\times 0.24 mol=0.24 mol of NaCl

1 mole = N_A=6.022\times 10^{23} molecules/ atoms

Number of NaCl molecules in 0.24 moles :

=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules

1.445\times 10^{23} formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

4 0
3 years ago
You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate
bekas [8.4K]

Answer : The volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid  (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

4.00=4.20+\log \left(\frac{(\frac{0.200x}{100})}{(\frac{0.100(100-x)}{100})}\right)

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid  (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

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What fact shows that baking a chocolate chip cookie is a chemical reaction?
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Answer:C The final product cannot be converted back to the original ingredients.

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