Amount of oxygen in the compound = 160 g
Amount of oxygen in the compound = 20.2 gm
Mole of oxygen in the compound = 160/16
= 10 moles
Mole of hydrogen in the compound = 20.2/1.01
= 20 moles
Then
The ratio of oxygen to ration of hydrogen = 1:2
So
The empirical formula of the compound is H2O. I hope the answer has come to your help.
Answer:
maybe they don't want to answer it only if you give them a lot of points and brainliest they will answer it i think
Explanation:
1 mole of any substance contains 6.022 × 1023 particles.
⚛ 6.022 × 1023 is known as the Avogadro Number or Avogadro Constant and is given the symbol NA
N = n × NA
· N = number of particles in the substance
· n = amount of substance in moles (mol)
· NA = Avogardro Number = 6.022 × 10^23 particles mol-1
For H2O we have:
2 H at 1.0 each = 2.0 amu
1 O at 16.0 each = 16.0 amu
Total for H2O = 18.0 amu, or grams/mole
It takes 18 grams of H2O to obtain 1 mole, or 6.02 x 1023 molecules of water. Think about that before we answer the question. We have 25.0 grams of water, so we have more than one mole of water molecules. To find the exact number, divide the available mass (25.0g) by the molar mass (18.0g/mole). Watch how the units work out. The grams cancel and moles moves to the top, leaving moles of water. [g/(g/mole) = moles].
Here we have 25.0 g/(18.0g/mole) = 1.39 moles water (3 sig figs).
Multiply 1.39 moles times the definition of a mole to arrive at the actual number of water molecules:
1.39 (moles water) * 6.02 x 1023 molecules water/(mole water) = 8.36 x 1023 molecules water.
That's slightly above Avogadro's number, which is what we expected. Keeping the units in the calculations is annoying, I know, but it helps guide the operations and if you wind up with the unit desired, there is a good chance you've done the problem correctly.
N = n × (6.022 × 10^23)
1 grams H2O is equal to 0.055508435061792 mol.
Then 23 g of H2O is 1.2767 mol
To calculate the number of particles, N, in a substance:
N = n × NA
N = 1.2767 × (6.022 × 10^23)
N= 176.26
N=
Answer:
<em>Sodium dihydrogen phosphate + calcium carbonate</em>
<u>Full ionic equation</u>
2 Na⁺(aq) + 2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)
<u>Net ionic equation</u>
2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)
<em>Sodium oxalate + calcium carbonate</em>
<u>Full ionic equation</u>
2 Na⁺(aq) + C₂O₄²⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + CaC₂O₄(s)
<u>Net ionic equation</u>
C₂O₄²⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + CaC₂O₄(s)
<em>Sodium hydrogen phosphate + calcium carbonate</em>
<u>Full ionic equation</u>
2 Na⁺(aq) + HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + 2 Na⁺(aq) + CO₃²⁻(aq)
<u>Net ionic equation</u>
HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + CO₃²⁻(aq)
Explanation:
Let's consider two kind of equations:
- Full ionic equation: includes all ions and species that do not dissociate in water.
- Net ionic equation: includes only ions that participate in the reaction (<em>not spectator ions</em>) and species that do not dissociate in water.
