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4 years ago

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

2 answers:

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]4 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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Milk of magnesia, a suspension of mg(oh)2 in water, reacts with stomach acid (hcl) in a neutralization reaction. mg(oh)2(s) + 2

defon

Balanced equation for the above reaction is as follows;
Mg(OH)₂ + 2HCl ---> MgCl₂ + 2H₂O
stoichiometry of Mg(OH)₂ to MgCl₂ is 1:1
mass of Mg(OH)₂ reacted - 1.82 g
number of moles of Mg(OH)₂ - 1.82 g/ 58.3 g/mol = 0.0312 mol
number of Mg(OH)₂ moles reacted - number of MgCl₂ moles formed
number of MgCl₂ moles formed - 0.0312 mol
mass of MgCl₂ formed - 0.0312 mol x 95.2 g/mol = 2.97 g
mass of MgCl₂ formed - 2.97 g

5 0

3 years ago

Is the boiling point of oxygen higher than nitrogen's

PSYCHO15rus [73]

The boiling point of oxygen is higher than nitrogen's boiling The reason the boiling point of O2 is higher is not because of increased van der Waals interactions, but simple physics. The mass of a molecule of O2 is greater than that of a molecule of N2, so the molecule of O2 traveling at a speed sufficient to break out of the liquid phase has a greater kinetic energy than an analogous N2 molecule. The net effect is that more energy must be distributed throughout a sample of O2 to achieve a given vapor pressure (in this case equal to atmospheric pressure) than for a sample of N2. More energy means greater temperature.

8 0

4 years ago

When a glucose molecule loses a hydrogen atom as the result of an oxidation-reduction reaction, the molecule becomes _____.

butalik [34]

Answer:

Explanation:

3 0

3 years ago

The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the con

gtnhenbr [62]

Answer:

Product A: cis; no

Product B: cis: no

Explanation:

Two common methods of forming oxiranes from alkenes are:

Reaction with peroxyacids
Formation of a halohydrin followed by reaction with base

1. Reaction with peroxyacids

(a) Stereochemistry

The reaction with a peroxyacid is a syn addition, so the product has the same stereochemistry as the alkene.

The starting alkene is cis, so the product is <em>cis</em>-2,3-diethyloxirane.

(b) Configuration

The product is optically inactive because it has an internal plane of symmetry.

It will not rotate the plane of polarized light.

2. Halohydrin formation

(a) Stereochemistry

The halogenation of the alkene proceeds via a cyclic halonium ion.

The backside displacement of halide ion by alkoxide is also stereospecific, so a cis alkene gives a cis epoxide.

The product is <em>cis</em>-2,3-diethyloxirane.

(b) Configuration

The cyclic halonium ion has an internal plane of symmetry, as does the product (meso).

The oxirane will not rotate the plane of polarized light.

8 0

3 years ago

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago