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pychu [463]
3 years ago
15

How many particles of Na are there in 1.43g of a molecular compound with a Molar mass of 23g?

Chemistry
1 answer:
Olin [163]3 years ago
7 0

Answer:

3.74 x 10²² particles

Explanation:

Given parameters:

Mass of compound  = 1.43g

Molar mass of compound  = 23g

Unknown:

Number of particles of sodium = ?

Solution:

To find the number of particles of Na in the compound, we need to obtain the mass of sodium from the total mass given;

          Mass of sodium  = \frac{molar mass of Na}{molar mass of compound} x mass of sample

                                      = \frac{23}{23}  x 1.43g

                                       = 1.43g

Now find the number of moles of this amount of Na in the sample;

          Number of moles  = \frac{mass}{molar mass} = \frac{1.43}{23}  = 0.062mole

Now;

                    1 mole of substance  = 6.02 x 10²³ particles

                       0.062 mole of substance  =  0.062 x 6.02 x 10²³ particles

                                                                     = 3.74 x 10²² particles

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Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = 62.364\text{ L.mmHg }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0
3 years ago
A 25 liter balloon holding 1.5 moles of carbon dioxide leaks. If we are able to determine that 0.9 moles of carbon dioxide escap
ad-work [718]

Answer:

10 Litre

Explanation:

Given that ::

v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole

Using the relation :

(n2 * v1) / n1 = (n2 * v2) / n2

v2 = (n2 * v1) / n1

v2 = (0.6 mole * 25 Litre) / 1.5 mole

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v2 = 10 Litre

3 0
3 years ago
Write the reaction of crystal violet with 1 equivalent of HCl.
tankabanditka [31]

The reaction of crystal violet with 1 equivalent of HCl will be when leuco crystal reacts with HCl to crystal violet it forms hexamethyl pararosaniline chloride.

<h3>What are crystal violet?</h3>

These are the trinary methane compounds mainly used for staining and dying of anything like bacteria or fungi and other name for it is methyl violet 10 B.

In reaction the HCl gets protonated and lone pair of nitrogen atom gives them positive charge.

3C6H6(3NH)3 + HCl will give 3C6H6(3NH)3NH4Cl + 2H

Therefore, reaction of crystal violet with 1 equivalent of HCl will be when leuco crystal reacts with HCl to crystal violet it forms hexamethyl pararosaniline chloride.

learn more about crystal violet, here:

brainly.com/question/12305949

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3 0
2 years ago
Which describes any substance that shatters or breaks easily?
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Answer:

Brittle

Explanation:

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Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

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K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
3 years ago
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