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user100 [1]
2 years ago
11

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

Chemistry
2 answers:
irina [24]2 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  = <u>    16   </u>

                                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]2 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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Answer is: 8568.71 of baking soda.

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V(H₂SO₄) = 17 L; volume of the sulfuric acid.

c(H₂SO₄) = 3.0 M, molarity of sulfuric acid.

n(H₂SO₄) = V(H₂SO₄) · c(H₂SO₄).

n(H₂SO₄) = 17 L · 3 mol/L.

n(H₂SO₄) = 51 mol; amount of sulfuric acid.

From balanced chemical reaction: n(H₂SO₄) : n(NaHCO₃) = 1 :2.

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n(NaHCO₃) = 102 mol, amount of baking soda.

m(NaHCO₃) = n(NaHCO₃) · M(NaHCO₃).

m(NaHCO₃) = 102 mol · 84.007 g/mol.

m(NaHCO₃) = 8568.714 g; mass of baking soda.

4 0
2 years ago
There are two types of nucleic acids, dna and rna. nearly all organisms use dna, not rna, as the central repository for genetic
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Found the choices. Pls see attachment. 

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A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f
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P1V1/T1 = P2V2/T2

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8 0
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