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NeX [460]
3 years ago
10

which tool was mostly likely used in a procedure if the lab report shows that approximately 300 ml of water was used

Chemistry
1 answer:
DiKsa [7]3 years ago
7 0

it is most likely a beaker, beacuse 300ml is quite a large volume. Otherwise, it would be a measuring cylinder or pippette

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shutvik [7]
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6 0
3 years ago
Read 2 more answers
Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
2 years ago
Calculate the mass percent (m/m) of a solution prepared by dissolving 45.09 g of NaCl in 174.9 g of H2O.
inysia [295]

Answer:

20.3 %  NaCl

Explanation:

Given data:

Mass of solute = 45.09 g

Mass of solvent = 174.9 g

Mass percent of solution = ?

Solution:

Mass of solution = 45.09 g + 174.9 g

Mass of solution = 220 g

The solute in 220 g is 45.09 g

220 g = 2.22 × 45.09

In 100 g solution amount of solute:

45.09 g/2.22 = 20.3 g

Thus m/m% = 20.3 %  NaCl

5 0
3 years ago
Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a
kvv77 [185]

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles

From the second titration, the moles of excess of HCl is:

n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g

The percentage of Al(OH)₃ is:

Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3%

3 0
3 years ago
Using the knowledge of kinetic particle theory, when salt is dissolved in a glass of water without stirring, all of the water so
Paha777 [63]

Explanation:

This is because the salt particles are colliding with the water particles

6 0
3 years ago
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