Answer:
14700J
Explanation:
From the question given, the following were obtained:
M = 100g
ΔT = 35° C
C = 4.2J/g °C
Q=?
The heat transferred can calculated for by using the following equation
Q = MCΔT
Q = 100 x 4.2 x 35
Q= 14700J
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9
Answer:
Inorganic chemistry: Inorganic chemistry is the study of chemicals that, in general, are not primarily based on carbon. Analytical chemistry: Analytical chemistry is the study of the composition of matter. It focuses on separating, identifying, and quantifying chemicals in samples of matter.
HOPE THIS HELPED!!!!!!!!!!XDDDDD
Answer:
The mean free path = 2.16*10^-6 m
Explanation:
<u>Given:</u>
Pressure of gas P = 100 kPa
Temperature T = 300 K
collision cross section, σ = 2.0*10^-20 m2
Boltzmann constant, k = 1.38*10^-23 J/K
<u>To determine:</u>
The mean free path, λ
<u>Calculation:</u>
The mean free path is related to the collision cross section by the following equation:

where n = number density

Substituting for P, k and T in equation (2) gives:

Next, substituting for n and σ in equation (1) gives:
