Answer:
100 / 500
Explanation:
100 / 500 is equivalent to 1 / 5.
We must first write out the entire equation for this reaction which is as follows:
CO + Fe2O3 --> Fe + CO2
Now we must balance this equation which provides us with the following equation:
3 CO + Fe2O3 --> 2Fe + 3 CO2
We are told that we have excess Fe2O3, so that suggests that CO is the limiting reagent. We now simply convert the mass of Fe given to moles of Fe, and convert moles of Fe to moles of CO.
35.0 g Fe/ 55.845 g/mol = 0.627 moles Fe
0.627 moles Fe x (3 moles CO)/(2 moles Fe) = 0.940 moles CO
Now with the moles of CO present, we simply convert this back to mass using the molecular weight of CO.
0.940 moles CO x 28.01 g/mol = 26.3 g CO.
Therefore, 26.3 g of CO are needed to produce 35.0 g of Fe. Since we began with three significant figures in our starting mass, our answer must also have three significant figures.
CO+2 H2=CH3OH
2.85 mol Co x (2mol H2/1 mol Co)=5.70 mol just concert to grams
5.70 mol H2 x (2 g H2/1 mol H2) =11.40 grams of H2
Answer:
Conduction is heat transfer through the direct contact of two subjects. The best example of this would be C, because it is the direct transfer of heat from the lightbulb to your hand. :)
A 0.50 M solution of a monoprotic acid HA with a pH of 2.24 would be, first, a weak acid, as it does not dissociate fully. This leaves us with an equilibrium expression: HA (aq) <span>⇌ H+ (aq) + A- (aq)
Where A- is the conjugate base of the weak acid.
In a study of equilibrium, we remember that the ka value is the acid dissociation constant, and has the equation:
Ka = (concentration of H+)(concentration of conjugate base)/concentration of acid
We know the concentration of H+ and A- are 10^-2.24 by the definition of a pH being the -log(concentration of H+).
The concentration of the acid has gone down a little bit, as it has partially dissociated into H+ and A-, so we'll have to subtract 10^-2.24 from 0.50 for the concentration of the acid to account for the dissociation.
The final equation would then become:
[H+]*[A-]/[HA] = Ka
(10^-2.24) * (10^-2.24) / (0.50 - 10^-2.24) = Ka
(3.31 * 10^-5) / (0.494) = Ka
Ka = 6.70 * 10^-5</span>