Answer: STP
CaCO3 = 5 g
Convert gram to mol
100 g of CaCO3 = 1 mol
5 g of CaCO3 (n) = 5 g *(1 mol/100 g) = 0.05 mol
Gas law
PV =nRT
V = nRT/P
V = (0.05 mol * (0.08206 L atm /K mol) *273 K)/1 atm
V = 1.124 L
Explanation:
Assuming that you mean table sugar (sucrose), then at room temperature and without any catalyst, there is no reaction.
However if you elevate and hold the temperature of the aqueous solution at 50 to 60 °C (especially in the presence of a suitable catalyst, like mineral acid) the sucrose dimer will split into glucose and fructose. This is called hydrolysis and the resulting solution is called an invert sugar solution.
The reaction could be written as:
C12H22O11 (sucrose) + H2O (water) → C6H12O6 (glucose) + C6H12O6 (fructose)
or
C12H22O11 (aq) + H2O (l) → C6H12O6 (aq) + C6H12O6 (aq)
Notice that both of the produced sugars have the same empirical formula. Check with your instructor or in your textbook to see if more exact formulas are needed.
10.861% / 100 = 0.10861
12.428 % / 100 = 0.12428
(0.10861 * 187.9122) + (0.12428 * 190.9407) <span>+ (0.76711 x 192.8938)</span>
= 192.1100 amu .
hope this helps!
D. this is because they consume less resources, but produce just as much power. if this is wrong, which i doubt, it is a.
Answer:
umm i believe that the one you chose is correct
Explanation: