Answer:
1.52V
Explanation:
Oxidation half equation:
2Al(s)−→2Al^3+(aq) + 6e
Reduction half equation
3Sn2^+(aq) + 6e−→3Sn(s)
E°cell= E°cathode - E°anode
E°cathode= −0.140 V
E°anode= −1.66 V
E°cell=-0.140-(-1.66)
E°cell= 1.52V
What intermolecular forces are present in each of the substances? CH4,C3H8,CH3F,HF, C6H5OH (dispersion forces<span>, </span>dipole-dipole forces<span>, or </span>hydrogen bonding<span>);A sample of ideal gas at room temperature occupies a volume of 34.0L at a pressure of 782torr .</span>
The mass of NaCl formed is 8.307 grams
<u><em> calculation</em></u>
step 1: write the equation for reaction
Na₂CO₃ + 2HCl → 2 NaCl +CO₂ +H₂O
Step 2: find the moles of Na₂CO₃
moles = mass/molar mass
The molar mass of Na₂CO₃ is = (23 x2) + 12 + ( 16 x3) = 106 g/mol
moles = 7.5 g/106 g/mol =0.071 moles
Step 3: use the mole ratio to determine the mole of NaCl
Na₂CO₃:NaCl is 1:2 therefore the moles of NaCl =0.07 x2 =0.142 moles
Step 4: calculate mass of NaCl
mass= moles x molar mass
the molar mass of NaCl= 23 +35.5 =58.5 g/mol
mass = 0.142 moles x 58.5 g/mol =8.307 grams
