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Alekssandra [29.7K]
3 years ago
13

How does water change as it freezes?

Chemistry
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

Hi there!

Your answer is:

A.

Explanation:

When frozen, water turns from a liquid to a solid! An example of this is a glass of water. You fill the glass with liquid tap water, and then put ice cubes in it. The ice cubes are solid and the tap water is liquid!

I hope this helps!

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NiS2(s) + O2(g) --> NiO(s) + SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical
makkiz [27]

Answer:

1. The theoretical yield of NiO is 5.09g.

2. O2 is the limiting reactant.

3. The percentage yield of NiO is 95.5%

Explanation:

Step 1:

The balanced equation for the reaction is given below:

2NiS2(s) + 5O2(g) —> 2NiO(s) + 4SO2(g)

Step 2:

Determination of the masses of NiS2 and O2 that reacted and the mass of NiO produced from the balanced equation. This is illustrated below below:

Molar mass of NiS2 = 59 + (32x2) = 123g/mol

Mass of NiS2 from the balanced equation = 2 x 123 = 246g

Molar mass of o3= 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar mass of NiO = 59 + 16 = 75g/mol

Mass of NiO from the balanced equation = 2 x 75 = 150g

Summary:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2 to produce 150g of NiO

Step 3:

Determination of the limiting reactant. This can be obtain as follow:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2.

Therefore, 11.2g of NiS2 will react with = (11.2 x 160)/246 = 7.28g of O2.

From the above calculation, we can see that it will take a higher mass of O2 i.e 7.28g than what was given i.e 5.43g to react completely with 11.2g of NiS2.

Therefore, O2 is the limiting reactant and NiS2 is the excess reactant.

1. Determination of the theoretical yield of NiO.

In this case, the limiting reactant will be used as all of it is consumed in the reaction. The limiting reactant is O2.

From the balanced equation above, 160g of O2 reacted to produce 150g of NiO.

Therefore, 5.43g of O2 will react to produce = (5.43 x 150)/160 = 5.09g of NiO.

Therefore, the theoretical yield of NiO is 5.09g.

2. The limiting reactant is O2. Please review step 3 above for explanation.

3. Determination of the percentage yield of NiO. This is illustrated below:

Actual yield of NiO = 4.86g

Theoretical yield of NiO = 5.09g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 4.86/5.09 x 100

Percentage yield of NiO = 95.5%

3 0
3 years ago
When scientists create a representation of a complex process, they are
Burka [1]

Answer:

d making models.

Explanation:

When scientists create a representation of a complex process, they are inferring that they are making models.

A model is an abstraction of the real world or a complex process. Models are very useful in developing solutions to processes that are not easily simplified.

  • The models allow a part of a body to be simply studied.
  • Through this simple abstraction, extrapolations to other parts of the system can be deduced.
  • This can give very useful insights into the other parts of the system.
  • The heterogeneity of complex processes is a huge limitation to understanding them.
  • A homogenous part can be modelled and used to understand the system.
5 0
3 years ago
If the temperature of the helium balloon were increased from 30°C to 35°C and the volume of the balloon only expanded from 0.47L
Andrej [43]

Answer:

R u from ridge?

Explanation:

Mr Wilson trippin

6 0
3 years ago
What is Charles's law?<br> State the definition of the law in words.
aev [14]

Answer:a law stating that the volume of an ideal gas at constant pressure is directly proportional to the absolute temperature

Explanation:

4 0
3 years ago
How many grams of sodium fluoride should be added to 300. mL of 0.0310 M of hydrofluoric acid to produce a buffer solution with
Kisachek [45]

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF = K_a=6.8\times 10^{-4}

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

pK_a=4-\log (6.8)

pK_a=3.17

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

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

pH=pK_a+\log \frac{[NaF]}{[HF]}

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

2.60=3.17+\log (\frac{[NaF]}{0.0310})

[NaF]=0.00834M

Now we have to calculate the moles of NaF.

\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole

Now we have to calculate the mass of NaF.

\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g

Therefore, the mass of sodium fluoride added should be 0.105 grams.

4 0
4 years ago
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