Answer:
0.011 moles
Explanation:
There are about 6.02*10^23 atoms in a mole, so in the given sample, there are
which is about 0.011 moles.
Which element requires the least amount of
energy to remove the most loosely held electron
from a gaseous atom in the ground state?
<h3>Answer-</h3><h3>Na</h3>
First its ... higher energy of reactants , higher energy of products & you do the same with the lower ones than you do transition state
<span>The question does not mention the brand or size of antacid tablets, but some research in the internet shows that for a very common brand, Alka Seltzer, that it takes about 4 tablets to produces 60 mL of gas. So for 120mL it will take 120/60 x 4 or 8 tablets to produce 120 mL of CO2 gas.</span>
Here we will use the general formula of Nernst equation:
Ecell = E°Cell - [(RT/nF)] *㏑Q
when E cell is cell potential at non - standard state conditions
E°Cell is standard state cell potential = - 0.87 V
and R is a constant = 8.314 J/mol K
and T is the temperature in Kelvin = 73 + 273 = 346 K
and F is Faraday's constant = 96485 C/mole
and n is the number of moles of electron transferred in the reaction=2
and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)
so by substitution :
0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q
∴ Q = 4.5 x 10^-26
