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3 years ago

How many H atoms in 12H2O?

Chemistry

1 answer:

Ilya [14]3 years ago

3 0

Answer:

24 atoms of H or 1.4 x 10²⁵ hydrogen atoms

Explanation:

simple method

1 H₂O has atoms of hydrogen and 1 atom of oxygen

while 12H₂O will have 24 atoms of hydrogen and 12 atom of oxygen

by Avagadros number

molar mass of water H₂0=18.01528 g/mol

1 mole of H₂0 have 2 moles of Hydrogen

one mole of water= 6.02⋅10²³water molecules =1.2 x 10²⁴hydrogen atoms

12 mole of H₂O = 1.2 x 10²⁴ x 12= 1.4 x 10²⁵ hydrogen atoms

1.4 x 10²⁵ hydrogen atoms in 12 moles of H₂O

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Which type of bond is formed by two atoms that equally share one pair of electrons? (1 point) ionic covalent metallic un?

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Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom

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Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

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Fifth isotope. =178.946amu

sixth isotope .179.947amu

Average atomic weight of Hafnium= sum of all the atomic weights of the isotopes/ Total occuring isotopes

Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu

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2 years ago

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

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Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

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