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3 years ago

Can Someone Help Me With This Please?

Chemistry

1 answer:

Ratling [72]3 years ago

8 0

Answer:

Explanation:

This is going to be a very d-u-m-b question, but when you touch the Mg rod, why doesn't it kill you? After all, it is giving away a lot of electrons, isn't it?

The reason it doesn't kill you is because after the electrons are given away, they go into the liquid on the left side.

The important equation for this part of the reaction is

Mg ==> Mg2+ + 2e-

So the Mg goes from a solid to an ion (Mg2+) which goes into the liquid on the left.

The Mg must loose weight!!!!

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What are the problems with studying the structure of the earth?

elena55 [62]

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3 years ago

Which fuel is used for nuclear power?<br> a. coal<br> b. oil<br> c. natural gas<br> d. uranium

vekshin1

Hello,

The fuel used for nuclear power is Uranium.

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3 years ago

Read 2 more answers

The lewis dot structure for CSBr2?

AfilCa [17]

Answer:

See attached picture.

Explanation:

Hello!

In this case, by considering that the carbon has four valence electrons as it is in group IVA, sulfur has six valence electrons as it is in group VIA and bromine has seven valence electrons as it is in group VIIA we infer that the carbon is the central atom in CSBr2 so one double bond between carbon and sulfur makes it complete the octet and two carbon-to-bromine bonds are formed in order to complete the octets of both carbon and bromine as shown on the attached picture.

Best regards!

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3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

4 years ago

If the ka of a monoprotic weak acid is 1.2 × 10-6, what is the ph of a 0.40 m solution of this acid?

frez [133]

First, we write out a balanced equation.
HA <--> H(+) + A(-)

Next, we create an ICE table

HA <--> H+ + A-
[]i 0.40M 0M 0M
Δ[] -x +x +x
[]f 0.40-x x x

Next, we write out the Ka expression.

Ka = [H+][A-]/[HA]

Ka = x*x/(0.40-x)

However, because Ka is less than 10^-3, we can assume the amount of dissociation is negligible. Thus,

Assume 0.40-x ≈ 0.40

Therefore, 1.2x10^-6 = x^2/0.40

Then we solve for the [H+] concentration, or x

$\sqrt{0.40(1.2*10^{-6})} =x$

x=6.93x10^-4

Next, to find pH we do

pH = -log[H+]

pH = -log[6.93x10^-4]

pH = 3.2

5 0

3 years ago