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Sidana [21]
3 years ago
10

How many molecules are in 100 g of C6H1206?

Chemistry
1 answer:
rjkz [21]3 years ago
8 0

Answer: 100 grams of glucose represents 0.55 moles .

Explanation:

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The air all around us is a mixture of gasses containing nitrogen, oxygen, carbon dioxide, argon
KiRa [710]

Answer:

Total pressure = 4.57 atm

Explanation:

Given data:

Partial pressure of nitrogen = 1.3 atm

Partial pressure of oxygen = 1824 mmHg

Partial pressure of carbon dioxide = 247 torr

Partial pressure of argon = 0.015 atm

Partial pressure of water vapor = 53.69 kpa

Total pressure = ?

Solution:

First of all we convert the units other into atm.

Partial pressure of oxygen = 1824 mmHg / 760 = 2.4 atm

Partial pressure of carbon dioxide = 247 torr / 760 = 0.325 atm

Partial pressure of water vapor = 53.69 kpa / 101 = 0.53 atm

Total pressure = Partial pressure of N +  Partial pressure of O +  Partial pressure of CO₂ +  Partial pressure of Ar +  Partial pressure of water vapor

Total pressure = 1.3 atm + 2.4 atm + 0.325 atm + 0.015 atm + 0.53 atm

Total pressure = 4.57 atm

4 0
3 years ago
Post laboratory reflections 1 write equation [2] in the net ionic form 2. describe what happens when strong acid acts on limesto
oee [108]
It deteriorates. Definitely more than when you started.

4 0
2 years ago
(select all that apply.) the elements most often found in organic molecules are _____.
Arturiano [62]
<span>C. Carbon. H. Hydrogen. N. Nitrogen. O. Oxygen. P. Phosphorus. <span>S. Sulfur.</span></span>
7 0
2 years ago
Read 2 more answers
Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4
Nataliya [291]

Answer:

(a) 10.62

(b) 2.82

(c) 1.95

Explanation:

The neutralization reaction in this question is

(CH3CH2)3N + HCl   ⇒ (CH3CH2)3NH⁺ + Cl⁻

The problem  can be  solved by calculating the number of moles of triethylamine after  addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.

The pOH for a buffer of a weak base is gven by

pOH = pKb + log [(CH3CH2)3NH⁺] / [(CH3CH2)3N]

(a) 11 mL of 0.100 M HCl

mol HCl = 0.011 L x 0.100 mol/L = 0.0011 mol HCl

mol  (CH3CH2)3N reacted = 0.0011 mol

mol (CH3CH2)3NH⁺ produced = 0.0011 mol

mol (CH3CH2)3N  initially = 0.020 L x 0.1000 mol/L 0.0020 mol

mol (CH3CH2)3N left = 0.0020 mol - 0.0011 = 0.0009 mol

pKb = - log Kb = - log (5.2 x 10⁻⁴) = 3.284

Now we can compute pOH,

pOH = 3.284 + log ( 0.0011 / 0.0009 ) = 3.37

pH = 14 - pOH = 14 - 3.37 = 10.62

(b) 20.60 mL HCl

mol HCl = 0.0206 L x 0.100 mol/L = 0.00206

mol  (CH3CH2)3N consumed = 0.0020 mol

This is so  because the acid will consume completely the 0.0020 mol of the weak base  we had originally present.

Now the problem circumscribes to that of calculating the pH of the unreacted HCl

Total Vol = 0.0206 L + 0.02 L = 0.0406 L

mol HCl = 0.0206 L x .100 = 0.00206 mol

mol HCl left = 0.00206 mol - 0.0020 mol = 0.00006 mol

[HCl] = 0.00006 mol / 0.0406 L = 0.0015 M

Since HCl is a strong acid ( 100 % ionization) :

pH = - log [H⁺] = - log ( 0.0015 ) = 2.82

(c) We will compute the pH in  the same way we did for part (b)

mol HCl = 0.025 L x 0.100 mol/L = 0.0025 mol

mol HCl left = 0.0025 mol  - 0.0020 mol = 0.0005

Total Volume = 0.020 L + 0.025 L = 0.045 L

[HCl] = 0.0005 mol / 0.045 L = 0.111

pH = - log ( 0.111) = 1.95                                            

3 0
3 years ago
Please help me with these! Mark brainliest
bearhunter [10]

Answer:

1. Alkalinity

2. True

3. Red

Explanation:

1. 8-14, which are the bases, are alkaline

2. Salt lowers the melting/freezing point of water

3. Blue --> Red is Acid

6 0
3 years ago
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