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Ostrovityanka [42]
2 years ago
12

Room temperature is about 20 degrees Celsius. Explain how you could convert this temperature to kelvin. Use evidence from the fi

gure to support your answer.
Chemistry
1 answer:
Taya2010 [7]2 years ago
8 0

Answer:

293.15 K.

Explanation:

It is given that, the room temperature is 20 degrees Celsius.

We need to convert this temperature into kelvin.

The conversion from degrees Celsius to Kelvin is as follows :

T_k=T_c+273.15

We have, T_c=20^{\circ} C

So,

T_k=20+273.15\\\\T_k=293.15\ K

So, the room temperature is 293.15 kelvin.

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We have an object with a density of 620 g/ cm3 and a volume of 75 cm3. What is the mass of this object?
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2 years ago
why is it less effective to wash an insoluble precipitate with 15 ml of water once than it is to wash the precipitate with 3 ml
MArishka [77]

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6 0
2 years ago
Calculate the mass of 25,000 molecules of nitrogen gas. (1 mole = 6.02 x 1023 molecules)
Ainat [17]

Hey there!

Molar mass N2 = 28.01 g/mol

Therefore:

28.01 g N2 -------------- 6.02*10²² molecules N2

( mass N2 ?? ) ----------- 25,000 molecules N2

mass N2 =  ( 25,000 * 28.01 ) /  ( 6.02*10²³ )

mass N2 = 700250 / 6.02*10²³

mass N2 = 1.163*10⁻¹⁸ g


Hope that helps!

7 0
2 years ago
Read 2 more answers
A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2
Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

  • The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, <em>ΔTf </em>is the depression in freezing point of water.

<em>i</em> is van't Hoff factor.

<em>Kf </em>is the molal depression constant.

<em>m</em> is the molality of the solute.

  • Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
  • van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

4 0
2 years ago
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