Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole
The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited
Mg3(PO4)2 - the molar mass would be 262g/mol, which is 100%
Atomic mass of Mg is 24, since we have 3Mg we multiply by 3 and get a mass of 72
262 : 100% = 72 : x%
x = 72*100 / 262
x = 27.5%
And do that for every element — get the molar mass of P and multiply by 2, use a ratio, and get the molar mass of O and multiply by 8 and use ratios :)
I think the answer is C. acidic
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)
<u>Answer: </u>The amount of heat released is 84 calories.
<u>Explanation:
</u>
The equation used to calculate the amount of heat released or absorbed, we use the equation:
where,
Q = heat gained or released = ? Cal
m = mass of the substance = 10g
c = specific heat of aluminium = 0.21 Cal/g ° C
Putting values in above equation, we get:
Q = -84 Calories
Hence, the amount of heat released is 84 calories.