Answer:
I never did this but i think its d 37.5٪ lmk if i got it right pls and srry if i didnt
Answer:
Explanation:
2 HCl(g) + Mg(s) → MgCl₂(s) + H₂(g)
Let's calculate the quantity of mole of produced hydrogen with the Ideal Gases Law
P . V = n . R .T
2.19 atm . 6.82L = n . 0.082 . 308K
(2.19 atm . 6.82L) / (0.082 . 308K) = n
0.591 mol = n
1 mol of H₂ gas came from 2 mol of hydrochloric, so, 0.591 mol came from the double of mole
0.591 .2 = 1.182 mole of acid.
Molar mass of HCl = 36.45 g/m
1.182 mole are (36.45 g/m . 1.182g ) contained in 43.1 g
Density HCl = HCl mass / HCl volume
0,118 g/mL = 43.1 g / HCl volume
43.1 g / 0.118 g/mL = 365.3 mL (HCl volume)
2 <span>KOH +1 H3AsO4 →1 K2HAsO4 + 2 H2O</span>
The compound that is formed is: MgI2
Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!
http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm
<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V
<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)
</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
</span>
Sr2+(aq) + 2 e– <span> Sr(s) V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)
<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span> </span><span> Fe(s) V= -0.44 V
</span>Cu -> C2+ V = - 0.337V
V= - 0.777V (no spontaneous)
