Answer:
The experimental feature of the MALDI-MS technique which allows the separation of ions formed after the adduction of tissue molecules:
B) Velocity of ions depends on the ion mass-to-charge ratio.
Explanation:
- The option a is not correct as distance traveled by ions doesn't depend upon the ion charge rather it depends upon time for which you leave the sample to run.
- The option b is correct as velocity of ions depends on the ion mass-to-charge ratio because separation is done due to mass to charge ratio feature.
- The option c is incorrect as time of travel is not inversely proportional to the ion-to-mass ratio because the ion will move across the gel until you stop the electric field.
- The option d is not correct as electric field between MALDI plate and MS analyzer is though uniform but this feature doesn't allow the separation of ions.
Answer:
Due to an electron-pair acceptor and donor.
Explanations:
<em><u>Lewis acid</u></em> can be defined as an electron-pair acceptor. An example is Hydrogen ion(H+). This is because it is a proton and it distributes positive charge which means that it accepts electrons(negative charge).
<em><u>Lewis base</u></em> can be defined as an electron-pair donor. This is because it donates electrons to be accepted by the proton. An example is ammonia(NH3).
Answer:
Mg3N2
Explanation:
it would be magnesium as it would loss to electron so it would have 10 electron. you can see in the picture above .
hope this helps :)
Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.
Answer:
B hope help you stay happy
