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2 years ago

7 Be + 2 HCl →BeCl2+ H2

Chemistry

1 answer:

Mashutka [201]2 years ago

3 0

Answer:

The percent yield is 42.52%.

Explanation:

Theoretical yield of beryllium chloride = 12.7 g

Actual yield of the beryllium chloride = 5.4 g

The formula used for the percent yield will be :

$\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Now put all the given values in this formula, we get:

$\%(yield)=\frac{5.4 g}{12.7 g}\times 100=42.52\%$

The percent yield is 42.52%.

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ASAP , 8.81 g Carbon

xz_007 [3.2K]

The empirical formula : C₂Cl₇

The molecular formula : C₁₀Cl₃₅

<h3>Further explanation</h3>

Given

8.81 g Carbon

91.2 g Chlorine

Molar Mass: 1362.5 g/mol

Required

The empirical formula and molecular formula

Solution

Mol ratio :

C = 8.81 g : 12.011 g/mol =0.733

Cl = 91.2 g : 35,453 g/mol = 2..572

Divide by 0.733

C : Cl = 1 : 3.5 = 2 : 7

The empirical formula : C₂Cl₇

(The empirical formula)n = the molecular formula

(C₂Cl₇)n = 1362.5

(2x12.011+7x35.453)n=1362.5

(272.193)n=1362.5

n = 5

6 0

3 years ago

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

8 0

3 years ago

What is SO2 shape name?

muminat

Answer:Molecular Formula SO2

Hybridization Type sp2

Bond Angle 119o

Geometry V-Shaped or Bent

Explanation:

hope this helped <3

5 0

3 years ago

PLEASE SOMEONE HELP ME WITH THIIS QUESTION

sdas [7]

Answer:

The correct answer is - 5 carbon compounds due to low to high intermolecular forces between their molecules.

Explanation:

Bottle C has gas in it and we know that alkane has carbon and hydrogen only which means they have a single sigma bond between them and very low intermolecular forces in between molecules and are present mostly at gaseous state. Thus, bottle C has alkane.

Alcohols have -OH group that can form rarely two pi bonds which means they have intermediate intermolecular force whereas acids have -cooH group with a high molecular force so bottle B with liquid is alcohol and A has acid.

3 0

3 years ago

What is kinetic energy? *

finlep [7]

Answer:

energy which a body possesses by virtue of being in motion

Explanation:

the work needed ro accelerate a body of given mass

4 0

2 years ago