b. a chemical that cannot be broken down or separated into other chemicals
Answer:
91 millilitres
Explanation:
Recommended application = 65mg / Kg
This means 65 mg of dicyclanil per kg (1 kg of body mass).
Concentration = 50 mg / mL
How many millilitres required to treat 70kg adult?
If 65mg = 1 kg
x = 70 mg
x = 70 * 65 = 4550 mg
Concentration = Mass / Volume
50 mg/mL = 4550 / volume
volume = 4550 / 50 = 91 mL
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
Wait wait wait, did you mistype your question? Water and water would just mix together, agar is thicker than water.
According to half life equation:
T(1/2) = ㏑2 / K1
when the T(1/2) = 4 min * 60 = 240 sec
by substitution:
240 = 0.6931 / K1
K1 = 2.9 x 10^-3
when the second T(1/2) = 20 sec, so to get K2:
T(1/2) = 0.6931 / K2
by substitution:
20 = 0.6931 / K2
∴K2 = 3.4 x 10^-2
so, we can get T2 by using this formula:
㏑ (K2/K1) = Ea/R (1/T1 - 1/T2)
by substitution:
㏑(3.4 x 10^-2)/(2.9 x 10^-3) = (24520 / 8.314) (1/298 - 1/T2)
∴ T2 = 396.7 K
= 396.7 - 273 = 123.7 °C
