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lianna [129]
2 years ago
8

How many sulfate ions are there in 321 grams of iron lll sulfate

Chemistry
1 answer:
nikitadnepr [17]2 years ago
7 0
Here, we use the mole as we would use any other collective number: a dozen eggs; a Bakers' dozen; a Botany Bay dozen.
Of course, the mole specifies a much larger quantity, and if I have a mole of stuff then I have
6.022
×
10
23
individual items of that stuff. We can also specify an equivalent mass, because we also know the mass of a mole of iron, and a mole of oxygen etc........The mole is thus the link between the macro world of grams and kilograms and litres, that which we can measure out in the lab, to the micro world of atoms, and molecules, that which we can perceive only indirectly.
Here we have the formula unit
F
e
2
(
S
O
4
)
3
. If there is a mole of formula units, there are necessarily 2 moles of iron atoms, 3 sulfate ions,.......etc.
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What volume of 12M HCI is needed to prepare 250<br> of 0.20M HCI?
Alchen [17]

Answer: 4.2

Explanation:

M_{A}V_{A}=M_{B}V_{B}\\(12)V_{A}=(250)(0.20)\\V_{A}=\frac{(250)(0.20)}{12}=\boxed{4.2}

6 0
2 years ago
In 1990, the northern spotted owl was listed as endangered. Because of this, millions of acres of forests in the Pacific
Gwar [14]

Answer:

the endangered spices act

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8 0
3 years ago
Enter your answer in the provided box.
nexus9112 [7]

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, V_1=43.1\ L

Initial temperature, T_1=24^{\circ} C=24+273=297\ K

Final temperature, T_2=78^{\circ} C=78+273=351\ K

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L

So, the final volume of the gas is 50.93 L.

4 0
3 years ago
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
3 years ago
How does the law of conservation of mass apply to this reaction C2H4+4O2--&gt;4H2O+2CO2 .Only the oxygen needs to be balanced B.
r-ruslan [8.4K]
C2H4+4O2-->4H2O+2CO2

the answer is C
6 0
3 years ago
Read 2 more answers
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