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VashaNatasha [74]
4 years ago
10

2. Which of the following is not a component of the plasma membrane

Chemistry
2 answers:
spayn [35]4 years ago
7 0
The answer is d because it said so
Sonja [21]4 years ago
7 0
The answer is D because every time it may consume something it may waste different acids and they don’t “waste” nucleic acids
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Why do chemist synthesize polymers in the lab?
Vesna [10]

Answer:Overview. Polymer chemists study large, complex molecules. They understand how the smaller building blocks (monomers) combine to form polymers, and they manipulate both their molecular structure and chemical or other processing to develop specific functional characteristics in an end product.

Explanation:

4 0
3 years ago
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A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?
Basile [38]

Answer:

The answer is 2.20 M

Explanation:

This is because ammonia has a pH of 11.8 and if you take 14-11.8 it equals 2.2 so the answer is 2.20 M

6 0
3 years ago
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A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H5CO2) solution with 0.1787 M KOH solution at 25°C. Calculate the p
natima [27]

Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of HC_2H_5CO_2 = 0.1917 M

Volume of HC_2H_5CO_2 = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of HC_2H_5CO_2

\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2\times \text{Volume of }HC_2H_5CO_2

\text{Moles of }HC_2H_5CO_2=0.1917M\times 0.220L=0.0422

As we known that at equivalent point, the moles of HC_2H_5CO_2 and KOH are equal.

So, Moles of KOH = Moles of HC_2H_5CO_2 = 0.0422 mol

Now we have to calculate the volume of KOH.

\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}

\text{Volume of }KOH=\frac{0.0422mol}{0.1787M}

\text{Volume of }KOH=0.00754

Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O

Moles of C_2H_5CO_2K = 0.0422 mol

\text{Concentration of }C_2H_5CO_2K=\frac{0.0422mol}{0.22754L}=0.1855M

At equivalent point,

pH=\frac{1}{2}[pK_w+pK_a+\log C]

Given:

pK_w=14\\\\pK_a=4.89\\\\C=0.1855M

Now put all the given values in the above expression, we get:

pH=\frac{1}{2}[14+4.89+\log (0.1855)]

pH=9.08

Therefore, the pH at equivalence is, 9.08

6 0
3 years ago
If you have 3.0 liters of an unknown gas, at standard temperature and pressure conditions, then you can calculate what about tha
Ivenika [448]
Density I'm pretty sure
5 0
4 years ago
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Which of the following has a central atom that exceeds the octet rule? PCl5 BeF2 H2S CO2
kherson [118]
Schwarzschild Radius here.

So,

The octet rule really only holds up for the second period elements, and only certain elements in that period, for that matter.  Elements in the third periods and beyond can have expanded octets.

Draw Lewis dot diagrams for each molecule.

PCl_5 :

P is the central atom bonded individually to each Cl.  The only logical diagram is that with three lone pairs surrounding each Cl and the P single-bonded to each Cl atom.  If you count the bonds, you will find that P has five pairs, or 10 electrons.  This exceeds the octet rule.

BeF_2 :

The Lewis dot diagram for this molecule is a linear molecule with the Be in the center, singly bonded to both F's, each of which have three lone pairs.  This means Be has only two pairs, or four electrons.  While not an example of a central atom that exceeds the octet rule, Be is one of the special elements that can stably have an incomplete octet.

H_{2}S :

The most stable diagram for this molecule is with S in the center with two lone pairs and singly bonded to each H atom.  With four pairs, or 8 electrons, this molecule obeys the octet rule.

CO_2 :

The diagram for this molecule is C in the center doubly bonded to each O atom.  Each O atom has 2 lone pairs.  C has four pairs, or 8 electrons.  This molecule also obeys the octet rule.

Hope this helps!
7 0
3 years ago
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