During ______ the composition of a substance does not change

Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50

OLga [1]

Answer:

$\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}$

Explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q = It

$\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}$

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

$\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}$

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.

(a) Calculate q

$\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}$

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

$\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}$

Note: The answer can have only one significant figure because that is all you gave for the time.

4 0

2 years ago

Because oxygen is more electronegative than hydrogen, the water molecule is

lapo4ka [179]

The water molecule has two hydrogen atoms joined to one oxygen atom by single covalent bonds. Because oxygen is more electronegative than hydrogen, the electrons of the covalent bond spend more time closer to the oxygen then hydrogen which creates a polar covalent bond. Hope this answers your question. Have a great day!

7 0

3 years ago

How does the radius of the central atom affect the boiling points of Group 16 halides?

Taya2010 [7]

I don’t really know but hope you figure it out

6 0

2 years ago

What net force must act on the book to make it accelerate to the left at a rate

Nata [24]

Answer:

Reach us now for support www.toplivewriters.com

Explanation:

3 0

2 years ago

(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

$m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2$

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

$m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu$

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

$m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3} \\\\m_{AgNO_3}=2.03gAgNO_3$

Best regards!

5 0

2 years ago