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3 years ago

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If the final mass of the system is 120g, what was the percent yield of the reaction?

1 answer:

Neko [114]3 years ago

7 0

Answer:

Sorry for the wrong answer plz

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Use the de Broglie's Wave Equation to find the wavelength of an electron moving at 7.3 × 106 m/s. Please show your work. Note: h

ad-work [718]

Answer:

$\lambda=9.96\times 10^{-11}\ m$

Explanation:

The expression for the deBroglie wavelength is:

$\lambda=\frac {h}{m\times v}$

Where,

$\lambda$ is the deBroglie wavelength

h is Planck's constant having value $6.62607\times 10^{-34}\ Js$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

v is the speed of electron.

Given that v = $7.3\times 10^6\ m/s$

Applying in the equation as:

$\lambda=\frac {h}{m\times v}$

$\lambda=\frac{6.62607\times 10^{-34}}{9.11\times 10^{-31}\times 7.3\times 10^6}\ m$

$\lambda=\frac{10^{-34}\times \:6.626}{10^{-25}\times \:66.503}\ m$

$\lambda=\frac{6.626}{10^9\times \:66.503}\ m$

$\lambda=9.96\times 10^{-11}\ m$

6 0

3 years ago

The volume occupied by 0.25 mol of sulfur dioxide at rtp <br>

Vladimir79 [104]

Answer:

V = 0.25 x 22.4 Liters

Explanation:

1 mole of any gas occupies 22.4 Liters at STP.

8 0

3 years ago

Compared to ultraviolet light, an electromagnetic wave that has a higher frequency will also have ________. (4 points)

Jet001 [13]

Answer: shorter wavelength and equal speed

5 0

3 years ago

What is the pH of 0.000134 M solution of HCI?

mafiozo [28]

Answer: The pH will be 3.87

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.000134$ moles of $HCl$ gives = $\frac{1}{1}\times 0.000134=0.0001342$ moles of $H^+$

Putting in the values:

$pH=-\log[0.000134]$

$pH=3.87$

Thus the pH will be 3.87

5 0

3 years ago

Read 2 more answers

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