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solmaris [256]
3 years ago
13

When practicing for your oral presentation, you should do all of the following except: Hold any notes or index cards low and bel

ow your waist. Speak very quickly to complete the presentation in a short time. Practice your presentation on your own or in front of a mirror. Practice your presentation in front of friends or family.
Chemistry
2 answers:
sergiy2304 [10]3 years ago
6 0
You don't want to speak very quickly, you want people to understand you. You shouldn't hold your cards at your waist either because your voice won't project to the class. It's best to practice in front of people to get used to it. 
rjkz [21]3 years ago
6 0

Answer:

Hold any notes or index cards low and below your waist.

Speak very quickly to complete the presentation in a short time.

Explanation:

When practicing for an oral presentation, it is best to practice in front of the people you are already familiar with, and who will give you valuable and sincere feedback. Families and friends will not shy away from giving you genuine feedback about your performance. Also, practicing in front of the mirror by oneself is also helpful, for it shows you how you are actually doing the presentation, and also reveals the things you need to change. You become the critic of yourself. Holding of index cards below the waist is also another thing to be avoided. Keep them above the waist, but not too high that they block your face. And, speaking fast is never a good thing while presenting anything. You want your audience to understand and know the things you speak of, not want them scrambling with the words you're throwing out in a rush, like a bullet train.

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A substance that cannot be chemically broken down into simpler substances is a an electron. b a heterogeneous mixture. c an elem
antiseptic1488 [7]

Answer:

c. an element.

Explanation:

An element -

It refers to the substance , which has same type of atoms , with exactly same number of protons , is referred to as an element .

In term of chemical species , elements are the smallest one , and can not be bifurcated down to any further small substance by the means of any chemical reaction .

Hence , from the given information of the question ,

The correct term is an element  .

3 0
3 years ago
For each described change, determine the generally expected impact on a salt's solubility. LABEL No solubility change, SOLUBILIT
Anna [14]

Answer:

A- Solubility decrease.

B- Solubility increase.

C- No solubility change.

Explanation:

Hello!

In this case, since the solubility of salt stands for the maximum amount of salt that can be added to a specific mass of water, usually 100 g; we need to take into account that for table salt in aqueous solution, the higher the temperature the larger the solubility and the lower the temperature the smaller the solubility; it means that more salt is dissolved in the same mass of water at higher temperatures and vice versa. Therefore, A- would decrease the solubility as the solution is cooled down and B- would increase the solubility as the solution is heated up.

Moreover, since the mass of water is assumed to remain the same, adding more salt do not affect the solubility but increase the degree of saturation of the solution up to supersaturated, yet the solubility remains unchanged.

Best regards!

4 0
3 years ago
What happens when iron is placed in a solution containing copper ions?
alexgriva [62]

When an iron is dipped in Copper Sulphate

Solution this reaction between them and

copper sulphate change into blue color to light

green color. This show that iron is more

reactive then copper, it can to replace copper

from CuSO4 , CuSO4 is of blue color and

FeSO4 is light green color.

Hope it helps

8 0
4 years ago
A reaction that occurs in the internal combustion engine is
Anna35 [415]

Answer:

Explanation:1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).

ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).

ΔrH = 180.6 kJ.

2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).

ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).

ΔS = 24.8 J/K.

3) ΔG = ΔH - TΔS.

55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.

2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.

3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.

7 0
3 years ago
The air in a 4.00 L tank has a pressure of 1.20 atm . What is the final pressure, in atmospheres, when the air is placed in tank
stellarik [79]

Answer:

P₂ = 1.92 atm

Explanation:

According to the boyle's law:

The pressure of given amount of gas at constant temperature and number of moles of gas is inversely proportional to the volume.

P₁V₁= P₂V₂

Given data:

Initial volume = 4.00 L

Initial pressure = 1.20 atm

Final volume = 2500  mL  

Final pressure = ?

Solution:

First of all we will convert the volume into litter.

2500 mL × 1L/1000 mL = 2.5 L

P₁V₁ =  P₂V₂

P₂ = P₁V₁ /V₂

1.20 atm×4.00 L = P₂ ×2.5 L

P₂ = 1.20 atm×4.00 L/ 2.5 L

P₂ = 4.8 atm. L/ 2.5 L

P₂ = 1.92 atm

4 0
3 years ago
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