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Viefleur [7K]
3 years ago
12

Determine the density of a liquid from the following data: Mass of the

Chemistry
1 answer:
Tanya [424]3 years ago
6 0

Answer:0.477 g/ml

Explanation:

Density=(40.14-33.79)/13.3 ml

Density=6.35/13.3

Density=0.477 g/ml

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During a chemical change items are lost or gained to make new substance true or false​
Otrada [13]

Answer:

If you meant Atoms then the answer is false

5 0
3 years ago
If a sample of air initially occupies 240L at 2 atm how much pressure is required to compress it to 20L at constant temperature
IceJOKER [234]

Answer:

24 atm.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 240 L

Initial pressure (P₁) = 2 atm

Final volume (V₂) = 20 L

Temperature = constant

Final pressure (P₂) =?

The final pressure required, can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

2 × 240 = P₂ × 20

480 = P₂ × 20

Divide both side by 20

P₂ = 480 / 20

P₂ = 24 atm

Thus, the final pressure required is 24 atm.

4 0
3 years ago
During the hot summer days, there is a lot more water in the air. This is due to which change of state?
Zarrin [17]
With the evaporation
5 0
3 years ago
For each item, choose whether the change is a physical change or chemical change
Aleonysh [2.5K]

Sorry, but where are the ‘items’?

8 0
3 years ago
68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi
Vera_Pavlovna [14]

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>

Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:

  • 0.528 mol Mg(NO₃)₂ * \frac{1molMg(OH)_2}{1molMg(NO_3)_2} = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

  • 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g
7 0
2 years ago
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