Which of the following can be true about a chemical reaction

A 3.50 amp power supply is used to deposit chromium from a solution of CrCl3. How long will it take to deposit 100.0 grams of ch

Norma-Jean [14]

Explanation:

The given data is as follows.

Current (I) = 3.50 amp, Mass deposited = 100.0 g

Molar mass of Cr = 52 g

It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.

Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

$z \times 52 g = 96500 \times 100 g$

z = $\frac{96500 \times 100 g}{52 g}$

= 185576.9 C

As we know that, Q = I × t

Hence, putting the given values into the above equation as follows.

Q = I × t

185576.9 C = $3.50 amp \times t$

t = 53021.9 sec

Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.

3 0

2 years ago

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago

The mass of an element is 27 and it's atomic number is 13. Give the composition of the nucleus of it's atom.

love history [14]

Answer:

nucleus consists of:

13 protons, 14 neutrons

8 0

2 years ago

What are "representative elements"?

Mars2501 [29]

Answer: c. elements in the same row of the periodic table

Explanation: It kind of represents a graph of the periodic table

8 0

2 years ago

Read 2 more answers

Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th

lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

"Your question is not complete, it seems to be missing the diagram of the emission spectrum"

the diagram of the emission spectrum has been added.

From the given chart;

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency of the wave
λ is the wavelength

$f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz$

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0

1 year ago