The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.
The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.
Thus the water sample of the chemists is both a mixture and pure substance.
Mass of Pd = 95.78 g which is close the the variant E) 95.89
Explanation:
To calculate the number of moles of palladium (Pd) we use the following formula:
number of moles = mass / atomic weight
mass = number of moles × atomic weight
mass of Pd = 0.90 × 106.42
mass of Pd = 95.78 g
Learn more about problems with number of moles:
brainly.com/question/3262
Answer:
No, IR should not soely be used to identify molecules
Explanation:
IR is a method that identifies the functional groups in a molecule by deducing the frequency of stretching and vibration of bonds. Each peculiar type of bond has a frequency for the vibration of each bond represented on the IR spectrum.
However, one method is never enough to identify a compound. A combination of methods must always be used to clear up ambiguities arising from overlapping IR frequencies. Also, interpretation of the nuanced peaks of the fingerprint region in IR spectra is quite challenging and only gives a fair idea of the functional groups present in the compound.
Therefore other methods such as NMR, UV-VISIBLE etc should also be involved in the identification of compounds.
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
<em />
Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
<em />
The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
<em />
The balance is A Go check other answer I posted
