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3 years ago

Given the following equation: 8 Fe + S8 —> 8 FeS what mass of iron is needed to react with 5.65 mol of sulfur S8

Chemistry

1 answer:

Mekhanik [1.2K]3 years ago

4 0

Answer:

2522g

Explanation:

8Fe + S8 —> 8FeS

From the question,

8moles of Fe required 1mole of S8.

Therefore, Xmol of Fe will require 5.65mol of S8 i.e

Xmol of Fe = 8 x 5.65 = 45.2moles

Now we need to covert 45.2moles to gram. This is illustrated below below:

Molar Mass of Fe = 55.8g/mol

Number of mole of Fe = 45.2moles

Mass of Fe =?

Mass = number of mole x molar Mass

Mass of Fe = 45.2 x 55.8

Mass of Fe = 2522g

Therefore, 2522g of Fe is needed for the reaction.

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

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3 years ago

Which choice lists the correct order of the coefficients of each substance in the following neutralization reaction when the equ

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3 years ago

What will be the pressure of 2.00 mol of an ideal gas at a temperature of 20.5 degrees Celsius and a volume of 62.3L?

erastova [34]

.774atm

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P being pressure, V being volume, n being moles, R being the gas constant, and T being temperature.

Before you start doing any of the math, make sure of two things. Since you're looking for pressure, you'll need a gas constant. When I did the problem, I used the gas constant of atm or atmospheres which is .0821.

Also! Remember to always convert celsius into kelvin, to do this, add 273 to the given celsius degree. After this is all set and done, your equation should look like this:

P = $\frac{2 x .0821 x 293.5}{62.3}$

The reason that the equation is divided by the volume is due to the fact that you need to isolate the variable or pressure.

Multiply everything on the top and divide by the bottom and you should receive the final answer of .774atm.

Hope this helps!

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