Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
Cathode rays are a stream of negatively charged particles or electrons in vacuum tubes. They were first seen by Johann Hittorf in 1869. They are named since they are emitted by electrode or cathode. In order to release the electrons, the electrons should first be detached from the atoms of the cathode.
58.7 %
Please correct me if I’m wrong. :)
Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
Answer:
Roman numbers are oxidation state of the metals
Because some elements of metals show more than one oxidation state like iron Fe2+ in ferrous and Fe3+ in ferric.
Explanation:
