What is the second most abundant gas in the atmosphere

Which of the following places the elements in the correct order of increasing first ionization energy?

Papessa [141]

A) Ca, Mg, Be
The first ionization energy increases as we move up a group because of the smaller atomic size and less shielding of electrons.

4 0

3 years ago

Read 2 more answers

Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but

alukav5142 [94]

Answer:

a) For nicotine, the protonated form is the present in stomach.

b) For caffeine, the neutral base form is the present in stomach.

c) For Strychinene, the protonated form is the present in stomach.

d) For quinine, the protonated form is the present in stomach.

Explanation:

In a basic dissociation for molecules with basic nitrogen, the equilibrium is:

A + H₂O ⇄ AH⁺ + OH⁻

Where A is neutral base and AH⁺ is protonated form

The basic dissociation constant, kb, is:

$K_{b} = \frac{[AH^+][OH^-]}{[A]}$

As pH in stomach is 2,5:

[OH] = $10^{-[14-pH]}$

[OH] = 3,16x10⁻¹² M

Thus:

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$ (1)

Using (1) it is possible to know if you have the neutral base or the protonated form, thus:

(a) nicotine Kb = 7x10^-7

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$221359 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For nicotine, the protonated form is the present in stomach

(b) caffeine,Kb= 4x10^-14

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$0,0127 = \frac{[AH^+]}{[A]}$

[AH⁺}<<<<[A]

For caffeine, the neutral base form is the present in stomach

(c) strychnine Kb= 1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$314456 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For Strychinene, the protonated form is the present in stomach

(d) quinine, Kb= 1.1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$347851= \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For quinine, the protonated form is the present in stomach.

I hope it helps!

7 0

3 years ago

As you are giving a friend directions to your house, you tell them that

natima [27]

The Rivergate because you are basing the distance off of there

8 0

3 years ago

Milk has a density of 1.04 g/ml. If you have a volume of 510 ml, what is the mass in grams

Law Incorporation [45]

Density is defined as the ratio of mass and volume. The formula of density is:

$density = \frac{mass}{volume}$ -(1)

Density of milk = $1.04 g/mL$ (given)

Volume of milk = $510 mL$ (given)

Substituting the values in formula (1):

$1.04 g/mL = \frac{mass}{510 mL}$

$mass= 1.04 g/mL \times 510 mL$

$mass = 530.4 g$

Hence, the mass of the milk is $530.4 g$ .

4 0

3 years ago

Photosynthesis allows plants to turn light energy into chemical energy by forming glucose from carbon dioxide and water: 6CO2(g)

dezoksy [38]

Answer:

305 g of CO₂

3.77 × 10⁵ kJ

Explanation:

Let's consider the global reaction for photosynthesis.

6 CO₂(g) + 6 H₂O(l) → C₆H₁₂O₆(g) + 6 O₂(g) ΔHrxn = 2802.8 kJ

A 1.70 lb sweet potato is approximately 73% water by mass. If the remaining mass is made up of carbohydrates derived from glucose (MW = 180.156 g/mol), how much carbon dioxide (MW = 44.01 g/mol) was needed to grow this sweet potato?

Let's consider the following relations:

The potato is 100%-73%=27% glucose by mass.
1 lb = 453.59 g.
6 moles of CO₂ produce 1 mole of glucose.
The molar mass of glucose is 180.156 g/mol.
The molar mass of carbon dioxide is 44.01 g/mol.

Then, for a 1.70 lb potato:

$1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{6molCO_{2}}{1molGlucose} .\frac{44.01gCO_{2}}{1molCO_{2}} =305gCO_{2}$

How much light energy does it take to grow the 1.70 lb. sweet potato if the efficiency of photosynthesis is 0.86%?

According to the enthalpy of the reaction, 2802.8 kJ are required to produce 1 mole of glucose. Then, for a 1.70 lb potato:

$1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{2802.8kJ}{1molGlucose} .\frac{1}{0.86\% } =3.77\times 10^{5} kJ$

7 0

3 years ago