The interaction between the two atoms of H in H2 with the lower energy corresponds to a covalent bond between hydrogen's.
When the two atoms of H form a bond, they are overlapping the individual orbitals to form a new one. Every hydrogen has 1 electron which sits in a 1s orbital and then form one molecular orbital. The energy of H2 is lower than individual hydrogens because the electrostatic interaction between them.
1.Calculate the number of mole of O2: 48/32=1.5 mol
2.Calculate the number of mole of H2 by mole ratio: 1.5 x 1/2=0.75 mol
3.Calculate the answer:
0.75 x 2=1.5g
Answer:
I know Z
Explanation:
because here is mixing two substance so there is mixture
Answer:
A) 0.1225 M
B) 100.4 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)
Step 2: Calculate the reacting moles of NaOH
17.73 mL of 0.1036 M NaOH react. The reacting moles are:
0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol
Step 3: Calculate the reacting moles of HA
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.
Step 4: Calculate the molar concentration of HA
1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:
M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M
Step 5: Calculate the molar mass of HA
1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:
0.1845 g / 1.837 × 10⁻³ mol = 100.4 g/mol
When electrical power is applied to two plates placed inside the water, electrolysis occurs and hydrogen appear on the cathode and Oxygen on anode. It is due to the ideal faradaic efficiency, the amount of hydrogen generated is twice the number of moles of oxygen.
If you look at the equation , reduction occurs at cathode i.e
2 H+ + 2e−<span> → H</span>2
Oxidation occurs at anode
2 H2O → O2 + 4 H+ + 4e<span>−
</span>
Overall reaction
2 H2O(l) → 2 H2(g) + O2(g<span>)
Therefore, </span><span> volume of gas collected over one electrode double the volume of gas collected over the other electrode.</span>
