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Lunna [17]
3 years ago
14

Calculate the missing variables in each experiment below using Avogadro’s law.

Chemistry
1 answer:
blagie [28]3 years ago
5 0

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml    n1 = 9.11 mol

V2 = 123 ml    n2 = ?

Formula

                               \frac{V1}{n1}  =  \frac{V2}{n2}

                                         n2 = \frac{n1V2}{V1}

                                         n2 = \frac{(9.11)((123)}{(612)}

                                                n2 = 1.83 mol                                                

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a scientist uses 68 grams of CaCo3 to prepare 1.5 liters of solution. what is the molarity of this solution?
victus00 [196]

Answer: 0.4533mol/L

Explanation:

Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol

68g of CaCO3 dissolves in 1.5L of solution.

Xg of CaCO3 will dissolve in 1L i.e

Xg of CaCO3 = 68/1.5 = 45.33g/L

Molarity = Mass conc.(g/L) / molar Mass

Molarity = 45.33/100 = 0.4533mol/L

7 0
3 years ago
What is the mass of water that results from combining 2.0 g of hydrogen with 16.0 g of oxygen?
astraxan [27]
The formula for water is H2O so there would have to be two Hyrdogens and one oxygen. Therefore it would be 4g of Hydrogen and 16g of Oxygen leaving you with 20g.
The answer is D.

Hope this helps :) ~
8 0
3 years ago
Could you please help
Semenov [28]
I would say D, because you need to start with nothing to measure the different sizes as they start to grow. hope this helps!
8 0
3 years ago
Find oxidation number for the aucl4 in haucl4 is it 1 or -1?
Romashka [77]
H^{+I}Au^{+VII}Cl_{4}^{-I}
7 0
3 years ago
How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 53.2 grams of hydrogen gas in the
olchik [2.2K]
The balanced chemical reaction is:

<span>2Na + 2H2O → 2NaOH + H2
</span><span>
We first use the amount of hydrogen gas to be produced and the molar mass of the hydrogen gas to determine the amount in moles to be produced. Then, we use the relation from the reaction to relate H2 to Na.

53.2 g H2 ( 1 mol / 2.02 g ) ( 2 mol Na / 1 mol H2 ) ( 22.99 g / 1 mol ) = 1210.96 g Na

1210.96 g Na ( 1 mL / 0.97 g ) = 1248.41 mL Na needed</span>
7 0
3 years ago
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