A. Tolerance range is different for different organisms.
Answer:
0.35 atm
Explanation:
To solve this problem, we use Boyle's Law: , where P is the pressure and V is the volume.
Here, V_1 = 0.355 L, P_1 = 1.0 atm, and V_2 = 0.125 L. So, just plug these values into the equation:
(1.0) * (0.355) = * (0.125) ⇒ ≈ 0.35 atm
Thus, the pressure is 0.35 atm.
Hope this helps!
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>