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Liula [17]
3 years ago
9

When 1000 C of charge is passed through CuSO4 solution, x g of copper is deposited. How much charge should be passed through the

solution to deposit 5x g of copper?​
Chemistry
1 answer:
Stels [109]3 years ago
6 0

Number of charge = 5018 C

<h3>Further explanation</h3>

Given

1000 C of charge for x grams of copper

Required

Number of charge

Solution

Faraday's Law :

\tt W=\dfrac{e.i.t}{96500}\\\\W=\dfrac{e.Q}{96500}

For 1000 C, W = x grams

\tt x=\dfrac{1000.e}{96500}=0.0104e

For 5x grams :

\tt 5x=\dfrac{e.Q}{96500}\\\\5\times 0.0104e=\dfrac{e.Q}{96500}\\\\Q=\dfrac{96500\times 5\times 0.0104e}{e}=5018~C

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in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co
kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5

mol of element G :

\tt mol=\dfrac{4}{16}=0.25

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0
3 years ago
If 4520 kj of heat is needed to boil a sample of water, what is the mass of water
Cerrena [4.2K]

Answer:

1,085g of water

Explanation:

If we have the value 4520kj is because the question is related to Energy and heat capacity. In this case, the law and equation that we use is the following:

                                                  Q= m*C*Δt  where;

Q in the heat, in this case: 4520kj

m is the mas

Δt= is the difference between final-initial temperature (change of temperature), in this exercise we don´t have temperatura change.

In order to determine the mass, I will have the same equation but finding m

                                          m= Q/C*Δt    without   m=Q/C

So: m= 4,520J/4.18J/g°C

      m= 1,0813 g

5 0
3 years ago
Find the mass of each of these substances? 2.40 mol NaOH
damaskus [11]
Na = 23 x 2.40 = 55.2
O = 16 x 2.40 = 38.4
H = 1 x 2.40 = 2.40

55.2 + 38.4 + 2.4 = 96

2.40 mol of NaOH = 96 amu

4 0
3 years ago
Which of the following chemical formulas represents a common acid?
wariber [46]
HCI is one of the most common acids out of the following
5 0
3 years ago
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
3 years ago
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