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prohojiy [21]
3 years ago
11

Calculate the molarity of the final solution of Erythrocin B (MW 879.9 g/mol) if a .5028g sample is dissolved in water diluted t

o 100 mL in a volumetric flask (Solution 1), and then a 5mL aliquot of the stock solution is diluted to 100mL using volumetric equipment. (solution 2)
Chemistry
1 answer:
Mashcka [7]3 years ago
7 0

Answer:

2.857 × 10⁻⁴ M

Explanation:

First, we will calculate the molarity of the solution 1.

M = mass of solute / molar mass of solute × liters of solution

M = 0.5028 g / 879.9 g/mol × 0.100 L

M = 5.714 × 10⁻³ M

We can find the concentration of the final solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 5.714 × 10⁻³ M × 5 mL / 100 mL

C₂ = 2.857 × 10⁻⁴ M

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<u>Answer:</u> The relation between the forward and reverse reaction is K_f=\frac{1}{K_r}

<u>Explanation:</u>

For the given chemical equation:

2NH_3(g)\rightleftharpoons N_2 (g)+3H_2(g)

The expression of equilibrium constant for above equation follows:

K_f=\frac{[N_2][H_2]^3}{[NH_3]^2}         .......(1)

The reverse equation follows:

N_2 (g)+3H_2(g)\rightleftharpoons 2NH_3(g)

The expression of equilibrium constant for above equation follows:

K_r=\frac{[NH_3]^2}{[N_2][H_2]^3}       .......(2)

Relation expression 1 and expression 2, we get:

K_f=\frac{1}{K_r}

Hence, the relation between the forward and reverse reaction is K_f=\frac{1}{K_r}

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3 years ago
Please help me this is my fourth attempt.
Vaselesa [24]

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

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7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

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Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

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The ionization energy equation for the given values follow:

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From the values of ionization energy above, it can be seen that the ionization energy increases as every succeeding electron is removed.

Second ionization energy is a little higher than the first one but there is a huge amount of difference between the third and second ionization energy.

This implies that the ion formed during second ionization energy has a stable configuration and it requires a humongous amount of heat to release the third electron.

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Answer:

A.

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