Answer:
The system is not at equilibrium and the reaction will proceed to the left.
Explanation:
Step 1: Write the balanced equation
H₂(g) + CO₂(g) ⇄ CO(g) + H₂O(g)
Step 2: Calculate the reaction quotient (Q)
The reaction is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.
Q = [CO] × [H₂O] / [H₂] × [CO₂]
Q = 0.610 × 0.695 / 0.425 × 0.500 = 2.00
Since Q ≠ Kc, the reaction is not at equilibrium.
Since Q > Kc, the reaction will proceed to the left.
The answer to this question it true
4. If you evaporate the water first, then the sand and salt will be nearly impossible to separate. That eliminates 1 and 2. You can't filter salt out from an aqueous solution, so that eliminates 3. 4 works because you can filter out the sand (which doesn't dissolve) then evaporate the water away from the salt.
Answer:
C₆H₆O₃
Explanation:
Calculation sequence:
% => grams => moles => reduce => empirical Ratio
Molecular multiple = Molecular Mass / Empirical Mass
C: => 57.1% => 57.1 g => 57.1/12 = 4.7583
H: => 4.8% => 4.8 g => 4.8/1 = 4.8000
O: => <u>38.1% => 38.1 g </u>=> 38.1/16 = 2.3813
TTL => 100% 100 g
Reduced Mole values =>
C : H : O => 4.7583/2.3813 : 4.8000/2.3813 : 2.3813/2.3813 => 2 : 2 : 1
∴ empirical formula => C₂H₂O
empirical formula weight => 2C + 2H + 1O = [2(12) + 2(1) + 1(16)] amu = 42 amu
molecular formula weight (given in problem) = 126 g/mole
The molecular formula is a whole number multiple of the empirical formula.
molecular multiple = 126 amu / 42 amu = 3
∴ molecular formula => (C₂H₂O)₃ => C₆H₆O₃
Answer:
0.116 g.
Explanation:
- Firstly, we can find the mass of C and H in the unknown compound:
mass of C = (no. of moles of C)(atomic mass of C) = (0.117 mol)(12.01 g/mol) = 1.405 g.
mass of H = (no. of moles of H)(atomic mass of H) = (0.233 mol)(1.01 g/mol) = 0.235 g.
∴ mass of O = mass of unknown sample - (mass of C + mass of H) = 3.50 g - (1.405 g + 0.235) = 1.86 g.
∴ no. of moles of O = (mass of O)/(atomic mass of O) = (1.86 g)/(16.0 g/mol) = 0.116 g.
