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mash [69]
3 years ago
11

What are properties ?

Chemistry
1 answer:
frozen [14]3 years ago
8 0

Answer:

a

Explanation:

In logic and philosophy, a property is a characteristic of an object; a red object is said to have the property of redness. The property may be considered a form of object in its own right, able to possess other properties.

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Why doesn't all soil look and feel the same? List 3 reasons
Katen [24]

Answer:

all soil are also different due to how they were form

3 0
2 years ago
What happens when The vapour obtained by dropping conc. H2SO4 in a mixture of KI and MnO2 is treated with hypo solution​
Shalnov [3]

Iodine is decolorized.

The first reaction stated in the question occurs as follows;

2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)

The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.

Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.

The equation of the titration reaction is;

2Na2S2O3 + I2→ Na2S4O6 + 2NaI

When this reaction takes place, iodine is decolorized due to its reduction to I^-.

6 0
2 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
Briefly explain how magnetic behavior is related to the arrangement of valence electrons in the outer orbitals
olga nikolaevna [1]

Answer:

Elements with unpaired valence electrons in the outer orbital are paramagnetic. Elements that are predicted to exhibit paramagnetic behavior may exhibit ferromagnetic behavior. Elements that contain only paired valence electrons are diamagnetic.

3 0
3 years ago
A 120.5 g sample of water loses -7525 J of energy. What was the
dusya [7]

Answer:14.93

Explanation:heat gained=heat lost

120.5 ×4.184×change=7525

7525÷504.172

Ans14.925

5 0
3 years ago
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