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Angelina_Jolie [31]
2 years ago
7

Which of the following is an example of a

Chemistry
1 answer:
Sergio039 [100]2 years ago
5 0

Answer:

dissolving sugar into water is a physical change

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A compound is 44.82% Potassium, 18.39% Sulfur, and 36.79% Oxygen. Write the empirical formula and name the compound.
sleet_krkn [62]

Answer:

Empirical formula = K₂SO₄

The name of the compound is Potassium tetraoxosulfate (vi)

Explanation:

Percentage composition of the elements:

Potassium = 44.82%

Sulfur = 18.39%

Oxygen = 36.79%

Mole ratio of the elements:

Potassium = 44.82/39 =1.15

Sulfur = 18.39/32 = 0.57

Oxygen = 36.79/16 = 2.29

Dividing by the smallest ratio in order to obtain the simplest ratio

Potassium = 1.15/0.57 =2

Sulfur = 0.57/0.57 = 1

Oxygen = 2.29/0.57 = 4

Therefore, the mole ratio of Potassium : Sulfur : Oxygen = 2:1:4

Empirical formula = K₂SO₄

The name of the compound is Potassium tetraoxosulfate (vi)

5 0
3 years ago
Anhydrous copper sulphate can be used in a test for water. What two things will happen when water is added to anhydrous copper s
Misha Larkins [42]

Answer:

thanks bro for helping me thanku so much

3 0
2 years ago
Which kind of energy does an object have because of its height above Earth's
jonny [76]

Answer:

C po

Explanation:

baka po pero pwd din ang B pero ok ang C

3 0
3 years ago
Read 2 more answers
When a barium atom loses two electrons what charge does it have
Darya [45]

Answer:

It becomes a positive ion and its radius decreases

Explanation:

As per the Octet rule, Barium has 2 electrons in its outermost shell. When it loses the two electron it gains two positive charge i.e Ba2+. As the barium loses the two electron from its outermost shell, the outermost shell becomes vacant and thus is no more considered as a part of atomic geometry of the barium atom and since the outermost shell is considered negligible the radius of barium atom reduces automatically.  

6 0
4 years ago
How much energy is used to melt 44.33 g of solid oxygen?
Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

4 0
3 years ago
Read 2 more answers
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