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Korvikt [17]
3 years ago
11

Which experimental feature of the MALDI-MS technique allows the separation of ions formed after the adduction of tissue molecule

s?
A) Distance travelled by ions depends on the ion charge.
B) Velocity of ions depends on the ion mass-to-charge ratio.
C) Time of travel is inversely proportional to the ion mass-to-charge ratio.
D) Electric field between the MALDI plate and the MS analyzer is uniform.
Chemistry
1 answer:
34kurt3 years ago
6 0

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.

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Complete the sentences to explain your reasoning.
vredina [299]

Answer:

1-b

2-weaker(option is incorrect)

3-a

Explanation:

1-b because iodine is more electronegative because of this negative on iodine will be more stable as negative charge on more electronegative element is more stable.

2-weaker as size of Te (Tellurium) is greater than S (sulphur) so bond length of H-Te is larger than H-S and therefore bond energy will be lesser and can easily give hydrogen in case of H-Te. as bond energy is inversly proportional to bond length.

3-a hydrogen has more negative electron affinity so hydrogen will have -1 charge and it will behave as a electron donar atom that is basic not acidic hence NaH is not acidic.

3 0
3 years ago
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
Grace [21]

Answer:

0.21 g

Explanation:

The equation of the reaction is;

NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)

Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles

Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles

Since the reaction is 1:1, NaCl is the limiting reactant.

1 mole of NaCl yields 1 mole of AgCl

0.00147 moles of NaCl yields 0.00147 moles of AgCl

Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol

= 0.21 g

3 0
2 years ago
Which of the following contains both ionic and covalent bonding?
djyliett [7]
I believe the correct answer from the choices listed above is the first option. The compound that contains both ionic and covalent bonding is KOH or potassium hydroxide. It contains one covalent<span> (O-H) and one that is </span>ionic<span> (K-O). Hope this helps.</span>
7 0
3 years ago
Read 2 more answers
Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is th
Hunter-Best [27]

Answer:

The answer is 0.36 kg/s NO

Explanation:

the chemical reaction of NH3 to NO is as follows:

4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)

We have the following data:

O2 Volume rate = 645 L/s

P = 0.88 atm

T = 195°C + 273 = 468 K

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we calculate the moles found in 645 L of O2:

P*V = n*R*T

n = P*V/R*T

n= (0.88 atm * 645L/s)/((0.08205 L*atm/K*mol) * 468 K) = 14.78 moles of O2

With the reaction we can calculate the number of moles of NO and with its molecular weight we will have the rate of NO:

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8 0
3 years ago
What volume does 18.0g of bromine occupy?
Varvara68 [4.7K]

Hey there!

Atomic mass Bromine ( Br ) = 79.9 u

Therefore:

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18.0 g Br --------------- volume ??

Volume Br = 18.0 * 22.4 / 79.9

Volume Br = 403.2 / 79.9

Volume Br = 5.046 L

hope this helps!

4 0
3 years ago
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