There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.
Explanation:
39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.
4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.
We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.
So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.
Learn more about:
Avogadro's number
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First solve the moles of oxgen present in the compound
mol O = 6.93 g O ( 1 mol O / 16 g O )
mol O = 0.43 mol H
then solve the moles of hydrogen present
mol H = ( 7.36 - 6.93) g H ( 1 mol H / 1 g H)
mol H = 0.43 mol H
so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
H2O2
Answer:
The products of self-ionization of water are OH⁻ and H⁺.
Explanation:
- The water is self ionized according to the equation:
<em>H₂O → OH⁻ + H⁺.
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The ionic product for water (Kw) = [OH⁻][H⁺] = 10⁻¹⁴.
Kw is also called "self-ionization constant" or "auto-ionization constant".
Answer:
Balanced reaction:
3 H2 (g) + N2 (g) → 2 NH3 (g)
Use stoichiometry to convert g of H2 to g of NH3. The process would be:
g H2 → mol H2 → mol NH3 → g NH3
12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3
Explanation: See above
Hope this helps, friend.
