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3 years ago

How many particles would be found in a 1.224 g sample of K2O

Chemistry

2 answers:

irina [24]3 years ago

6 0

Answer:

7.8286×10²¹ particles.

Explanation:

First we need to calculate the total molar mass of the compound, in this case:

Potassium (K) = 39.1 g/mol × 2 = 78.2 +

Oxigen (O) = 16 g/mol × 1 = 16

Total(K₂O) = 94.2 g/mol

Then, we calculate the number of moles of the compound in the sample, this is done dividing de mass of the sample by the molar mass:

$mol =\frac{1.224 g}{94.2 g/mol}$

mol = 0.013 moles in our sample.

Finally, we calculate the total number of particles. The costant known as Avogadro number (6.022×10²³) is the number of particles or atoms contained in a mole of any substance. We need to multiply the number of moles by the Avogadro number.

particles = 0.013 mol × (6.022×10²³ particles/mol) = 7.8286×10²¹ particles.

lbvjy [14]3 years ago

4 0

Answer:

9.96*10^21

Explanation:

Molar mass of K2O=29*2+16

= 74g per mol

number of moles in the sample= 1.224/ 74

=0.1654

Number of particles in 1 mole=6.0221409*10^23

Number of particles= 0.01654*6.0221409*10^23

=9.96*10^21

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The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

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$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

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