You would know that the variable is quantitative if it shows any number to express the quantity. For example, quantitative variables are 50°C, 5 atm, 2 moles, 100 L and so on. A variable is qualitative if it expresses a relative quantity but not expressing a number. Examples would be: few, too hot, several, or even describing the characteristics of a variable. Hence, when the variable is in grams, then that would be quantitative.
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
Answer:
0.052mL
Explanation:
1mole of a gas occupy 22.4L.
Therefore, 1 mole of CO2 will also occupy 22.4L.
If 1mole of CO2 occupies 22.4L,
Then 2.3moles of CO2 will occupy = 2.3 x 22.4 = 51.52L
coverting this volume to mL, we simply divide by 1000 as shown below:
51.52/1000 = 0.05152mL = 0.052mL
Given mass of Scandium = 50.0 g
Increase in temperature of the metal when heated = 
Heat absorbed by Scandium = 
The equation showing the relationship between heat, mass, specific heat and temperature change:
Where Q is heat =
m is mass = 50.0 g
ΔT =
On plugging in the values and solving for C(specific heat) we get,
=50.0g(C)()
C = 0.491
Specific heat of the metal = 0.491
An alkoxide is an organic functional group formed when a hydrogen atom is removed from a hydroxyl group of alcohol when reacted with a metal. It is the conjugate base of alcoho.
