In an electrochemical cell, electrons travel in which direction? * (a) from the anode to the cathode through the external circui

In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>= 
=  = 0.45 mol

Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.

According to the ideal gas equation, we have PV = nRT
V = .
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
In that case, V = 

= 34.8 l

Answer: 3.28 mins

Explanation:

Here’s how it breaks down:

Conversions

74 mph = 33.08 m/s

6.5 km = 6500 m

(6500 m)/(33.08 m/s) = 196.5 seconds

196.5 seconds is equivalent to 3.28 minutes

The solution to your inquiry yields P = 17.73 atm. Explanation: The volume V is 250 ml, equivalent to 0.25 liters (L), with a mass of 3.4 g and a temperature of 45°C, which converts to 318°K. We utilize the ideal gas law PV = nRT for the calculations.

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and cut (q) down to a third

Explanation:

According to the principle of Le Chatelier, when a system reaches equilibrium and a change is introduced, the system will respond to counteract that change.

Since P and Q are reactants, raising the amount of either one or both without a proportional rise in R (which is a product) will cause the equilibrium to move towards the right. Similarly, if R decreases while P and Q remain constant, this too will push the equilibrium to the right. Thus, Increase(P), Increase(q), and Decrease(R) will lead to a rightward shift in the equilibrium.

Conversely, raising R without increasing P and Q will draw the equilibrium to the left. Likewise, cutting down P and/or Q without a similar reduction in R will shift the equilibrium leftward. Therefore, Increase(R), Decrease(P), Decrease(q), and triple both (Q) and (R) will shift the equilibrium to the left.

If there are equivalent changes in P and Q, with R remaining unchanged, then the equilibrium remains stationary. So, tripling (P) while reducing (q) to one third will not alter the equilibrium.