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qaws [65]
3 years ago
10

If one body is positively charged and another one is negatively charged free electrons tend to ?

Chemistry
2 answers:
nordsb [41]3 years ago
8 0
Free electrons tend to move from the negatively charged body to the positively charged body. Static electricity or shock may also occur
lord [1]3 years ago
5 0

Answer is: Move from the negatively charged body to the positively charged body.

The electron (symbol: e⁻) is a subatomic particle whose electric charge is negative one elementary charge.

The proton (p⁺) is subatomic particle with a positive electric charge of +1e elementary charge.

Opposite charges (positive and negative) attract one another.  

The negatively charged body has extra electrons, more electrons than protons.

The positively charged body has less electrons than protons.

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HELP PLEASE THE OTHER 'ANSWER' ISNT EVEN AN ANSWER!
hodyreva [135]

Answer:

most likely that (2) the replicated experiment was performed incorrectly.

Why, u ask? u dare question me:

1- The initial experiment invalidness cannot be proven.

2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>

3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.

4- Already knowing the data and errors would increase the precision of the replicated experiment.

5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.

happy learning!

4 0
3 years ago
In 1865 a pond was surrounded by open fields today the same area is swampy and surrounded by a forest which process is responsib
Nookie1986 [14]

Answer:

erosion

Explanation:

3 0
3 years ago
Read 2 more answers
What happened to the wind speeds of the storms?
oksano4ka [1.4K]
They would most likely speed up.
6 0
3 years ago
If 10.57 g of magnesium reacts completely with 6.96 g of oxygen, what is the percent by mass of oxygen in magnesium oxide? Round
Deffense [45]

Answer:

39.7 %

Explanation:

magnesium + oxygen ⟶ magnesium oxide

   10.57 g         6.96 g               17.53 g

According to the <em>Law of Conservation of Mass</em>, the mass of the product must equal the total mass of the reactants.

Mass of MgO = 10.57 + 6.96

Mass of MgO = 17.53 g

The formula for mass percent is

% by mass = Mass of component/Total mass × 100 %

In this case,

% O = mass of O/mass of MgO × 100 %

Mass of O = 6.96 g

Mass of MgO = 17.53 g

% O = 6.96/17.53 × 100

% O = 0.3970 × 100

% O = 39.7 %

5 0
3 years ago
What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:
Levart [38]

Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.

A) Reaction with NaI :

Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .

The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)

NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.

1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane

The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)

B) Reaction with AgNO3 :

Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.

AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )

The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.

7 0
3 years ago
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