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Andreas93 [3]
3 years ago
14

Potrzebuje pomocy pilnie!! daję naj

Chemistry
1 answer:
Ymorist [56]3 years ago
8 0

I dont speacka you languadge

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How many square centimeters are in an area of 9.90 inch2 ?
USPshnik [31]

Answer:

63.8708 hope this helps :)

Explanation:

7 0
2 years ago
Which class of organic compounds contains nitrogen?
Westkost [7]
<span>(3) amina ........................</span>
6 0
3 years ago
Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass
yan [13]

Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)

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

\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles

\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol    

According to stoichiometry :

2 moles of NaCl are formed from = 2 moles of Na

Thus 16 moles of NaCl are formed from=\frac{2}{2}\times 16=16moles  of Na

Mass of Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g

Thus 368 grams of sodium reacted.

4 0
3 years ago
How many grams of iron can be made with 21.5g of Fe2O3
SIZIF [17.4K]

The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g

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

2Fe₂O₃ -> 4Fe + 3O₂

  • Molar mass of Fe₂O₃ = 159.7 g/mol
  • Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
  • Molar mass of Fe = 55.85 g/mol
  • Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

<h3>How to determine the mass of iron, Fe produced</h3>

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

Therefore,

21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe

Thus, 15.04 g of Fe were produced.

Learn more about stoichiometry:

brainly.com/question/9526265

#SPJ1

5 0
1 year ago
If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial
jeka94

Answer:

the  initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

KSCN \rightarrow K^ + SCN^-

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5  m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the  initial concentration of SCN- in the mixture is 0.00588 M

5 0
3 years ago
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