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salantis [7]
2 years ago
7

In the Northern Hemisphere, recall that atmospheric CO2 levels are highest in early spring and lowest during early fall. What ac

counts for this seasonal fluctuation in atmospheric CO2 levels? Select all that apply.
Chemistry
1 answer:
asambeis [7]2 years ago
4 0

Answer:

Photosynthesis

Explanation:

Photosynthesis can be defined as the process whereby plants, some protistans, and some bacteria utilize the energy from sunlight to produce glucose from water and carbon dioxide. In early spring, there is the shift of plants from scanty winter branches to abundant spring leaves. When the leaves on the trees reduced in the fall, dead plants break down throughout the winter due to the activities of microbes. The microbes decompose the plant materials and generate carbon dioxide. This leads to increase in the level of carbon dioxide in the atmosphere. In the spring, leaves return to the plants/trees and the rate of photosynthesis increases greatly. This consumes more carbon dioxide and the level of carbon dioxide in the atmosphere reduced.

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Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
2 years ago
The density of benzene at 15 ∘C is 0.8787 g/mL. Calculate the mass of 0.1200 L of benzene at this temperature.
Pavel [41]
Use the density to convert volume into mass. 

since the density is in g/ml and the volume was given in Liters, we need to first convert the Liters into mililiters. just multiply by 1000 or move the decimal three times.

0.1200 Liters= 120.0 mL

120.0 mL (0.8787 grams/ 1 mL)= 105 grams
3 0
2 years ago
Read 2 more answers
Balance the following equations:
lesya [120]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

(I) N_{2} + 3H_{2} --> 2NH_{3}

(II) P_{4}  + 5O_{2}  --> 2P_{2} O_{5}

(III) 2NaF + Br_{2}  -->2NaBr + F_{2}

(IV) 2ZnS + 3O_{2} --> 2ZnO + 2SO_2

(V) Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

7 0
2 years ago
A naturally occurring oil co-distills with water to produce an oil/water distillate that is 20% oil by weight. If the molecular
Gelneren [198K]

Answer:

Explanation:

Partial pressure of oil = mole fraction of oil x total pressure

mole fraction of oil = mole of oil / mole of water + mole of oil

= mole of oil = mass of oil / molecular weight of oil

= 20 / 100 = .2

mole of water = 80 / 18

= 4.444

mole fraction of oil =  .2 / .2 + 4.444

= .2 / 4.644

Partial pressure of oil = mole fraction of oil x total pressure

= (.2 / 4.644 ) x 760 mm

= 32.73 mm Hg .

3 0
3 years ago
Drag each tile to the correct box.
77julia77 [94]

this is my attachment answer hope it's helpful to you

3 0
3 years ago
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