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Paladinen [302]
3 years ago
15

How many grams of CaCl, are needed to prepare 120 mL of 450 M Cach solution?

Chemistry
1 answer:
harkovskaia [24]3 years ago
5 0

Answer:

<em>gggyyhhhhjjjjjjhhhhhhhhhhhhhjjjjj</em>

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<span>Permeable means that a material is full of tiny, connected airspaces that allows water to seep through it.</span>
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2.<br> An alkane has at least on C=C<br> bond.<br> ut of<br> Select one:<br> O True<br> O False
garik1379 [7]
I think it’s false?????
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4 years ago
The following two compounds each exhibit two heteroatoms (one nitrogen atom and one oxygen atom). In compound A, the lone pair o
Galina-37 [17]

Answer:

The answer is the photo attached

Explanation:

3 0
3 years ago
At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubil
sertanlavr [38]

Answer: Thus the solubility of N_2 gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of N_{2} in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,  C=k_{H}P.

k_{H}= Henry’s law constant = ?

The partial pressure (P) of N_{2} in water = 4.07 atm

\C= k_{H}\times P\\95.7mg=k_{H}\times 4.07

k_{H}=23.5

At pressure of 10.0 atm

C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg

Thus the solubility of N_2 gas in water, at the same temperature, is 235mg/100g

6 0
3 years ago
Iron is extracted from iron oxide in the Blast Furnace: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2
arsen [322]

a. mass of iron = 69.92 g

b. percent yield = 93%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

a.

Reaction

Fe₂O₃+3CO⇒2Fe+3CO₂

MW Fe₂O₃ :  159.69 g/mol

mol Fe₂O₃

\tt \dfrac{100}{159,69}=0.626

mol Fe₂O₃ : mol Fe = 1 : 2

mol Fe :

\tt \dfrac{2}{1}\times 0.626=1.252

mass of Fe(Ar=55.845 g/mol) :

\tt 1.252\times 55.845=69.92~g

b.

actual yield = 65 g

theoretical yield = 69.92 g

percent yield :

\tt =\dfrac{65}{69.92}=0.93=93\%

8 0
3 years ago
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