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andreev551 [17]
3 years ago
10

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3 At this temp, 0.3000 mol of H2 and 0.300 mol o

f I2 were placed in a 1L container to react. What concentration of HI is present at equilibrium? H2+I2<-> 2HI
Chemistry
1 answer:
Vlada [557]3 years ago
4 0
Your answer is: Kc = [HI]² / {[H2] [I2]} 
<span>53.3 = (2x)² / {(0.400M - x)(0.400M - x)} </span>
<span>sqrt(53.3) = 2x / (0.400M - x) </span>
<span>2.92 - 7.30x = 2x </span>
<span>x = 0.314M </span>

<span>[HI] = 2x = 0.628M

:) hopefully this helped. (:</span>
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NaOH + HCl → NaCl + H2O. What mass of water forms when 2.75 × 10–4 mol NaOH reacts completely? given the equation, 2.75 × 10–4 m
Hoochie [10]
The balanced chemical reaction is written as:

<span>NaOH + HCl → NaCl + H2O

We are given the amount of sodium hydroxide to be used up in the reaction. This will be the starting point for the calculation. 

2.75 x 10^-4 mol NaOH ( 1 mol H2O / 1 mol NaOH ) ( 18.02 g H2O / 1 mol H2O ) = 4.96 x 10^-3 g H2O</span>
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Plants are capable of using nitrogen directly from the atmosphere. True false
Nataly_w [17]
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How many atoms of N are in 0.82 g of NaNO3? ( molar conversion)
VMariaS [17]

The number of atoms N = 5.8 x 10²¹

<h3>Further explanation</h3>

A mole is a unit of many particles (atoms, molecules, ions) where 1 mole is the number of particles contained in a substance that is the same amount as many atoms in 12 gr C-12  

1 mole = 6.02.10²³ particles  

mass of N in 0.82 g of NaNO₃ (MW NaNO₃: 85 g/mol) :

\tt mass~N=\dfrac{Ar~N}{MW~NaNO_3}\times mass~NaNO_3\\\\mass~N=\dfrac{14}{85}\times 0.82\\\\mass~N=0.135~g

moles of N :

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The number of atoms N :

\tt N=n\times No\\\\N=0.0096\times 6.02\times 10^{23}\\\\N=5.8\times 10^{21}~atoms

4 0
3 years ago
Do anyone know the answer?
siniylev [52]

Answer:

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Explanation:

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