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2 years ago

Find the number of grams 4.00 moles of CU(CN)2

Chemistry

1 answer:

USPshnik [31]2 years ago

4 0

Answer:

462g

Explanation:

First, let us calculate the molar mass of Cu(CN)2. This is illustrated below:

Molar Mass of Cu(CN)2 = 63.5 + 2(12+14) = 63.5 + 2(26) = 63.5 + 52 = 115.5g/mol

Number of mole of Cu(CN)2 given from the question = 4moles

Mass = number of mole x molar Mass

Mass of Cu(CN)2 = 4 x 115.5

Mass of Cu(CN)2 = 462g

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Create an element card that has the following information:

Shtirlitz [24]

Answer:

Atomic number:6

Atomic mass:12.011

Chemical symbol:C

Name of element: Carbon, Carbone (french)

Image is in my explanation

Explanation:

6 0

3 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

What would the total mass of a 4.9 mole sample of Barium (Ba) be if its Molar Mass is 137g?

Aleksandr-060686 [28]

Answer:

m = 671 grams

Explanation:

Given that,

No of moles, n = 4.9

Molar mass of Barium, M = 137 g

Mass divided by molar mass is equal to no of moles. It can be given by the formula as follows :

$n=\dfrac{m}{M}\\\\m=n\times M\\\\m=4.9\times 137\\\\m=671.3\ g$

m = 671 grams

So, the total mass of the sample of Barium is 671 grams.

8 0

3 years ago

Where is the mass of an atom found? Explain.

muminat

Explanation:

Over 99.9% percent of an atom's mass resides in the nucleus. The protons and neutrons in the center of the atom are about 2000 times heavier than the electrons orbiting around it.

Because the electrons are so light by comparison,they represent only a tiny fraction of a percent of the atom's total weight.

Hope this answer helps you!

4 0

3 years ago

Find the number of grams 4.00 moles of CU(CN)2 ​

Find the number of grams 4.00 moles of CU(CN)2