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3 years ago

Wich value gives the number of particles in 1 mol of a substance

Chemistry

2 answers:

juin [17]3 years ago

5 0

Answer: 6.022x10²³ (Avogadro's Number)

Explanation: The number of particles in 1 mole of a substance is determined by the Avogadro's Number.

exis [7]3 years ago

4 0

6.02x10 the power of 23 ( Avogadro formula)

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A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

Andrews [41]

The given question is incomplete. The complete question is as follows.

A certain liquid has a normal boiling point of $124.2^{o}C$ and a boiling point elevation constant $k_{b} = 0.62 ^{o}C kg mol^{-1}$ . A solution is prepared by dissolving some sodium chloride (NaCl) in 6.50 g of X. This solution boils at $127.4^{o}C$ . Calculate the mass of NaCl that was dissolved. Round your answer to significant digits.

Explanation:

As per the colligative property, the elevation in boiling point will be as follows.

T = boiling point of the solution =

$T_{o}$ = boiling point of the pure solvent = $124.2^{o}C$

$K_{b}$ = elevation of boiling constant = $0.62 ^{o}C kg mol^{-1}$

We will calculate the molality as follows.

molality = $\frac{\text{maas of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in g}}$

i = vant hoff's factor

As NaCl is soluble in water and dissociates into sodium and chlorine ions so i = 2.

Putting the given values into the above formula as follows.

$T - T_{o} = i \times K_{b} \times \text{molality of solution}$

$(127.4 - 124.2)^{o}C = 2 \times 0.62 \times \frac{m}{60} \times \frac{1000}{650}$

m = 100 g

Therefore, we can conclude that 100 g of NaCl was dissolved.

3 0

3 years ago

A student wonders whether a piece of jewelry is made of pure silver. She determines that its mass is 3.25 g. Then she drops it i

Anettt [7]

Answer:

The jewelry is 2896.54_Kg/m^3 less dense than pure silver

Explanation:

Density of jewellery = (mass of jewellery) ÷ (volume of jewellery)

=3.25g ÷ 0.428mL = 0.00325Kg÷0.000000428m^3 = 7583.46Kg/m^3

The density of silver is 10490_Kg/m^3 which is (10490 - 7583.46) 2896.54_Kg/m^3 more dense than the jewellery

The density of Silver [Ag]

The weight of Silver per cubic centimeter is 10.49 grams or the weight of silver per cubic meter is 10490 kilograms, that is the density of silver is 10490 kg/m³; at 20°C (68°F or 293.15K) at a pressure of one atmospheres.

3 0

3 years ago

Why does the boiling point of a liquid vary with atmospheric pressure?

tigry1 [53]

Atmospheric pressure decreases because air is less dense at higher altitudes. Because the atmospheric pressure is lower, the vapour pressure of the liquid needs to be lower to reach boiling point.

8 0

3 years ago

Please help me I’m not sure with that

Alex73 [517]

What I can’t see the pic ?hhbc639l73773 ghcbhjbch

5 0

3 years ago

Read 2 more answers

Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure

Galina-37 [17]

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

where,

$\DeltaT_b$ = change in freezing point

$k_b$ = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of $KCl$ will be,

$KCl\rightarrow K^++Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M $KCl(aq)$ at standard pressure, the freezing point of 1.0 M $CaCl_2(aq)$ at standard pressure is lower.

4 0

3 years ago