All categories

2 years ago

Why do copper ions go to the negative electrode. Please help someone

Chemistry

2 answers:

Artemon [7]2 years ago

8 0

Because copper ions have a positive charge and are therefore attracted to the negative electrode

shepuryov [24]2 years ago

7 0

Because a copper ion looses electrons, meaning it's negatively charged, and positives and negatives attract.

You might be interested in

If you use 65.34 g of Al2S3, how many grams of AICl3 can be produced?

Doss [256]

Answer:

sorry po

indi ko po maintindihan

6 0

2 years ago

Arrange the following compounds from left to right in order of increasing percentage by mass of hydrogen: a) H2O, b) C12H26, c)

Nuetrik [128]

Answer:

(A) N4H6 (B) H2O (C) LiH (D) C12H26

Explanation:

The given compounds have been arranged from left to right in order of increasing percentage by mass of hydrogen.

The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.

4 0

2 years ago

4NH3 + 5O2-> 4NO + 6H2O How many molesof NO are formed if 824g of NH3 react ?

ZanzabumX [31]

824 g NH3 (1 mol/17 g NH3) (4 mol NO/4mol NH3)

48.47 moles NO

4 0

2 years ago

The heat of fusion for water is the amount of energy that must be absorbed for water to _____.

ivanzaharov [21]

I would say A.evaporate because the water is getting absorbed.

4 0

2 years ago

The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63x10^-3 at 527 C. Calculate the equilibrium partial

Studentka2010 [4]

Answer: The partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

Explanation:

We are given:

$K_c=4.63\times 10^{-3}$

$p_{COCl_2}=0.760atm$

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $4.63\times 10^{-3}$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $527^oC=527+273=800K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304$

The chemical reaction for the decomposition of phosgene follows the equation:

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

At t = 0 0.760 0 0

At $t=t_{eq}$ 0.760-x x x

The expression for $K_p$ for the given reaction follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

We are given:

$K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x$

Putting values in above equation, we get:

$0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656$

Negative value of 'x' is neglected because partial pressure cannot be negative.

So, the partial pressure for the components at equilibrium are:

$p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm$

Hence, the partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

6 0

2 years ago