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Len [333]
3 years ago
9

How many moles is 135 L of ammonia gas at STP

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
3 0

Answer:

i

Explanation:

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ILL MARK BRAINLIEST :)
DanielleElmas [232]

Explanation:

According to the law of conservation of mass, mass can neither be created nor destroyed but it can simply be transformed from one form to another.

For example, Na^{+} + Cl^{-} \rightarrow NaCl

Mass of Na = 23 g/mol

Mass of Cl = 35.5 g/mol

Sum of mass of reactants = mass of Na + mass of Cl

= 23 + 35.5 g/mol

= 58.5 g/mol

Mass of product formed is as follows.

Mass of NaCl = mass of Na + mass of Cl

= (23 g/mol + 35.5) g/mol

= 58.5 g/mol

As mass reacted is equal to the amount of mass formed. This shows that mass is conserved.

As a result, law of conservation of mass is obeyed.

3 0
3 years ago
A ___________________ is a combination of two or more atoms that are held together by covalent bonds.
Andre45 [30]
The answer is Molecule
3 0
3 years ago
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How might a renewable energy source become non-renewable???
poizon [28]

if it is used continuously

6 0
3 years ago
Whatis the SI unit used by chemists to measure the amount of an element or substance
Harman [31]
The answer is Mole, but send me a message if you need the equation for how to get it
3 0
3 years ago
Read 2 more answers
consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of
MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

\frac{44}{56}

= 0.785 g of CO2

So,

<u>14.4 g of CaO</u><u> </u>must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

<u>Needed = 11.31 g</u>

<u>Available CO2  = 13.8 g</u><u> </u>(given)

So CO2 is in excess , hence<u> CaO is the limiting reagent and product will produce from 14.4 g of CaO</u>

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g  of CaO produce =

\frac{100}{56}

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

\frac{Actual\ yield}{Theoritical\ yield}\times 100

\frac{19.4}{25.71}\times 100

= 75.44 %

6 0
3 years ago
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