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I am Lyosha [343]
3 years ago
14

-The picture goes to Activity 2

Chemistry
2 answers:
Alborosie3 years ago
4 0

Answer:

Activity 1: Since the only forces apposing the upward movement of the load are gravity and air resistance the formulas of which can both be solved with weight then with constant variables for gravity and air, effort IS directly proportional to weight.

Activity 2:

Inclined plane.

to make lifting easier; raising or lowering a load.

Explanation:

hope this helps.

sergey [27]3 years ago
3 0

Answer:

Ramp.

Explanation:

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Advantages of modern periodic table​
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Answer:

It organizes all of the elements and it allows you to easily find the groups of elements such as

Group 1 of the periodic table consists of hydrogen and the alkali metals

Group 2 consists of the alkaline metals (batteries!)

Groups 3–12 contain transition metals

Groups 13–16 each contain at least one metalloid

Group 17 contains halogens

Group 18 consists of noble gases which are stable.

It is easy to read and theres a song to help you memorize them online.

Explanation:

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3 years ago
Glycine has pKa values of 2.34 and 9.60. Therefore, glycine will exist predominately as a zwitterion in an aqueous solution with
san4es73 [151]

Answer:

Option (B) pH > 2.34 and pH < 9.60.

Explanation:

In a zwitterionic structure, overall negative charge should be 0.

It means, for glycine, the two functional groups e.g. -NH_{2} and -COOH should bear equal and opposite charges in it's zwitterionic structure.

pK_{a} of -COOH (in glycine) = 2.34

Therefore, at a pH > 2.34, -COOH group of glycine exists predominantly as -COO^{-}.

pK_{a} of  -NH_{2}  (in glycine) = 9.60

Therefore, at a pH < 9.60,  -NH_{2}  group of glycine exists predominantly as -NH_{3}^{+}.

Hence we can say that at 2.34 < pH < 9.60, glycine exists predominantly in zwitterionic form due to presence of -COO^{-} and -NH_{3}^{+} in equal amount.

3 0
3 years ago
Flammable and combustible materials should be stored according to
avanturin [10]
There fire label i think
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3 years ago
Calculate E o cell for the reaction: AgCl(s) + NO(g) → Ag(s) + Cl-(aq) + NO3-(aq) (in acidic solution) Use the fact that the red
kaheart [24]

Answer:

-0.74

Explanation:

Let's consider the following redox reaction.

AgCl(s) + NO(g) → Ag(s) + Cl⁻(aq) + NO₃⁻(aq)

We can identify the oxidation and reduction half-reactions.

Reduction (cathode): 1 e⁻ + AgCl(s) → Ag(s) + Cl⁻(aq)

Oxidation (anode): 2 H₂O(l) + NO(g) → NO₃⁻(aq) + 4H⁺(aq) + 3 e⁻

The standard cell potential (E°) is equal to the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red,cat - Ered,an = 0.22 V - 0.96 V = -0.74 V

7 0
3 years ago
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