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3 years ago

Determine the number of atoms of O in 60.1 moles of Fe₂(SO₃)₃.

Chemistry

1 answer:

kvasek [131]3 years ago

4 0

Answer:

3.310308*10^26

Explanation:

nO=9nFe2(SO3)3=9*60.1=540.9 moles

number of atoms: 540.9*6.02*10^23

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Select all of the following statements that represent the differences between a voltaic cell and an electrolytic cell. Group of

Sunny_sXe [5.5K]

Answer:

The electrodes will change in mass for only the electrolytic cell

Electrode with the lowest reduction potential is reduced in an electrolytic cell

A potential is generated when the voltaic cell runs

6 0

3 years ago

I really need help with this before 5:10

gtnhenbr [62]

CAN YOU HELP ME ANWSER MINE ILL HELP YOU

4 0

3 years ago

If 16 moles of al react with 3 moles of S8 how many moles of Al2 S3 will be formed

Gnom [1K]

Answer:

8 moles

Explanation:

Al reacts with $S_8$ to produce $Al_2S_3$ as

$Al+S_8\rightarrow Al_2S_3$

The balanced chemical equation is

$16Al+3S_8\rightarrow 8Al_2S_3$

In the reaction, 16 moles of $Al$ react with 3 moles of $S_8$ to produce 8 moles of $Al_2S_3$ .

8 0

3 years ago

How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0

3 years ago

CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of

neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0

3 years ago