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Elodia [21]
2 years ago
13

What mass in grams would 5.7L of hydrogen gas occupy at STP?​

Chemistry
1 answer:
tekilochka [14]2 years ago
8 0

Answer:  The correct answer is:  " 0.54 g " .

__________________________________________

Explanation:

Note that "hydrogen gas" is:  

H₂ (g)  ;   that is:  a "diatomic element" (diatomic gas) ;

_________________________________________

The molecular weight of "H" is:  1.00794 g ;

   (From the Periodic Table of Elements).

So, the molecular weight of:  H₂ (g)  is:

    " 1.00794 g * 2  = 2.01588 g ; {use calculator) ;

_________________________________________

Note the conversion for a gas at STP:

______

  1 mol of a gas = 22.4 L gas;

___

i.e. " 1 mol / 22.4 L " ;

____

So:     " 5.7 L H₂ (g)  *  \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;

The "L" ("literes" cancel out to "1" ;  since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

____

and we are left with:

____

 [5.7 * 2.104588 g ] / 22.4  =  ?  g ;

______________________

→ [ 11.9961516  g ] / 22.4 =

          0.53554248214  g ;l

_____________________________

We round this value to:  " 0.54 g " ;

 → since "5.7 L " has 2 (two) significant figures;  

     22.4 is an exact number conversion;

     and "5.7 L" has fewer significant figures than:

    " 2.104588 " ; or:  " 1.00794 " .

  → as such: We round to "2 (two) significant figures."

______________________________

Hope this is helpful.  Wishing you the best in your academic endeavors!

_______________________________

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Consider the following reaction:
Katyanochek1 [597]

Answer:

The three statements are true

Explanation:

For the reaction:

I₂O₅(s) + 5CO(g) → I₂(s) + 5CO₂(g)

State oxidation of iodine in I₂O₅ is:

5 O²⁻ = 10⁻

As you have 2 I and the molecule has no charge, <em>oxidation state of I is +5</em>.

The carbon in CO has an oxidation state of +2 and in CO₂ is +4. That means <em>the carbon is oxidized</em>

<em />

An oxidizing agent is a substance that produce the oxidation of the agent that reacts with this one. CO is oxidized because of I₂O₅ is producing its oxidation being <em>the oxidizing agent</em>

<em></em>

Thus,<em> the three statements are true</em>.

3 0
3 years ago
Why is the mass of the third subatomic particle ignored?
dmitriy555 [2]

Answer: The mass of electrons is mostly ignored because electrons are extremely small compared to neutrons and protons.

Explanation: A proton is about 1,836 times the size of an electron.

On the periodic table, the atomic number for each element can be found. This number is found by measuring the weight of 6.02 x 10^23 atoms of the element in grams. Electrons aren't ignored when finding exact math, but for the sake of simplification high school teachers will generally have you only count the number of protons and neutrons when calculating the mass of atoms.

3 0
1 year ago
Give examples which indicate that nylon fibres are very strong.​
Readme [11.4K]

Explanation:

is used for making ropes, used for climbing rocks and for making parachutes. Their usage shows that nylon fibres have high tensile strength

3 0
2 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
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Solve and show work. Li2S + 2 HNO3 --&gt; 2 LiNO3 + H2S (a) Calculate the mass of lithium sulfide that will react with 250 mL of
Elenna [48]

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Li S + H2 N2 O5 -> Li N2 O5 + H2 S

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Li^2  S^2  +  H^4 N^4 O^10  --> Li^2 N^4  O^10  +  H^4 S^2

7 0
2 years ago
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