Answer:
the water concentration at equilibrium is
⇒ [ H2O(g) ] = 0.0510 mol/L
Explanation:
- CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)
∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30
⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L
⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
replacing in Kc:
⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30
⇒ 0.0721 [ H2O(g) ] = 3.679 E-3
⇒ [ H2O(g) ] = 0.0510 mol/L
Answer:
Y Q W Z X
Explanations:
The most reactive element is the element that will displace an element from it compound . The most reactive element will replace the less reactive element in it compound.
Q+ + Y Reaction occurs
Since the reaction occurs the element Y which is more reactive displaced element Q from it compound.
Q+W+ Reaction occurs
The reaction occurs, that means element Q replaces element w from it compound. Element Q is therefore more reactive than element W.
W+Z+ Reaction occurs
The reaction also occurs . This is an indication that element W replaces element Z in it compound. This means element W is very reactive than element Z.
X+Z+ No reaction
There is no reaction here. This is an indication that element X is less reactive than element Z. This is why element X can't displace element Y in it compound.
It’s 65% average that your answer
1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.
2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.
3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.
4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.
5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
Balanced:
1. <span>Na2O + H2O ---> 2NaOH
2. </span><span>K2O + H2O ---> 2KOH
3. </span><span>MgO + H2O ---> Mg(OH)2
4. </span><span>CaO + H2O ---> Ca(OH)2
5. </span><span>SO2 + H2O ⇄ H2SO3
6. </span>SO3 + H2O ---> H2SO4
All except by 2 were balanced.
