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VashaNatasha [74]
3 years ago
12

Calculate the number of atoms in a 7.45x10^3 g sample of sodium

Chemistry
1 answer:
Whitepunk [10]3 years ago
7 0

The molecular mass of sodium = 23g


One mole of sodium consists of 23 g of sodium


One mole consists of 1 Avagadro(6.022 x 10^23 molecules) number of molecules

Therefore 23 g consists of 6.022 x 10^23 molecules of Sodium.

Hence 7.45 x 10^3 g of Sodium will consist of x no of molecules of sodium.

x no of molecules =

(7.45 x 10^3 g/23 g) x (6.022 x 10^23)

=1950.6 x 10^ 23 molecules (atoms).

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Answer:

This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the molesof X that are needed to 

produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:Mole ratios:In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.

Calculations:In such a way, since 21.00 moles of D are given, we need the mole ratio of D to X in order to get the answer, which according to the reaction is 3:4 based on their coefficients in the reaction. Hence, we calculate the required as follows:

Explanation:

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2 years ago
What molecules can be made using hydrogen, carbon, and nitrogen?​
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Ammonia

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8 0
3 years ago
A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex
Anestetic [448]

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

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m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

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