The correct equation
for the overall reaction can simply be obtained by adding the two separate
equations together. Now when you add the two equations together, the overall K can
be calculated by multiplying the individual K values. Therefore:<span>
K(overall) = K1 * K2 </span>
K(overall) = (1.6 x
10^-10) * (1.5 x 10^7)
<span>K(overall) = 2.4 x
10^-3</span>
Answer:
1.4 mols
4th answer
Explanation:
22. 5 g of O2 in moles = (22.5/32) mols = 0.703 mol
The stoichiometry between O2 and H2O =1: 2
Therefore H2O produced = 2 * 0.703 mols=1.406 mols
Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
Answer:
3.10×10¯⁵ ft³.
Explanation:
The following data were obtained from the question:
Density (D) of lead = 11.4 g/cm³
Mass (m) of lead = 10 g
Volume (V) of lead =.?
Density (D) = mass (m) / Volume (V)
D = m/V
11.4 = 10 / V
Cross multiply
11.4 × V = 10
Divide both side by 11.4
V = 10 / 11.4
V = 0.877 cm³
Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:
1 cm³ = 3.531×10¯⁵ ft³
Therefore,
0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³
0.877 cm³ = 3.10×10¯⁵ ft³
Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.
Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.
