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3 years ago

When an aqueous solution of silver nitrate is

Chemistry

1 answer:

Andre45 [30]3 years ago

8 0

Answer:

$K^{+}$ and $NO_{3}^{-}$

Explanation:

The equation for the reaction is AgNO3(aq) + KCl(aq) ==> AgCl(s) + KNO3(aq)

With all the ions, it is

$Ag^{+}$ (aq) + $NO_{3} ^{-}$ (aq) + $K^{+}$ (aq) + $Cl^{-}$ (aq) ==> AgCl(s) +

$K^{+}$ and $NO_{3} ^{-}$ do not change, so they are the spectator ions and are removed

The ionic equation is:

$Ag^{+}$ (aq) + $Cl^{-}$ (aq) ==> AgCl(s)

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Brainliest grind of 2014i-

aksik [14]

Answer:

IM just too pogchamp

Explanation:

5 0

3 years ago

What is the pH of the solution at the endpoint of a titration

Arada [10]

Answer:

The pH at the equivalence point is 7.0.

Explanation:

This is because the solution only contains water and a salt that is neutral.

Since neither H+ or OH-, molecules remain in the solution.

We can conclude that the equivalence point is 7.0.

5 0

3 years ago

We would like to estimate how quickly the non-uniformities in gas composition in the alveoli are damped out. Consider an alveolu

Vera_Pavlovna [14]

Answer: It will take 11.775 seconds.

Explanation: As a sphere with a diameter of 0.1 mm, the area of an alveolus is

A = 4.π.r²

r for an alveolus would be: r = 0.00005m or r = 5. $10^{-5}$ m

Finding the area:

A = 4.3.14.(5. $10^{-5}$ )²

A = 3.14. $10^{-8}$ m²

The concentration change is to be 90% of the final, so

c = 0.9.3.14. $10^{-8}$

c = 28.26. $10^{-9}$

The oxygen diffusivity is 2.4. $10^{-9}$ m²/s, that means in 1 second 2.4. $10^{-9}$ of oxygen spread in one alveolus area. So:

1 second = 2.4. $10^{-9}$ m²

t seconds = 28.26. $10^{-9}$ m²

t = $\frac{28.26.10^{-9} }{2.4.10^{-9} }$

t = 11.775s

For a concentration change at the center to be 90%, it will take 11.775s.

4 0

3 years ago

How much heat is necessary to raise the temperature of a 15.0 g metal from 15.0 C to 25.0 C if the specific heat of that metal i

Bumek [7]

The solution would be like this for this specific problem:

E=mcΔT 
= (15 g)(1.91 Jg∘C)(25 ∘C−15 ∘C)
= 28.65 * 10
= 286.5

I hope this helps and if you have any further questions, please don’t hesitate to ask again.

5 0

3 years ago

Need help ASAP!!!! what is the value (angle) for the C=C=O bond in Ketene i.e. CH2=C=O

stepan [7]

Answer:

$180^\circ$ by the VSEPR theory.

Explanation:

This question is asking for the bond angle of the $\rm C=C=O$ bond in $\rm H_2C=C=O$ . The VSEPR (valence shell electron pair repulsion) theory could help. Start by considering: how many electron domains are there on the carbon atom between these two bond?

Note that "electron domains" refer to covalent bonds and lone pairs collectively.

Each nonbonding pair (lone pair) of valence electrons counts as one electron domain.
Each covalent bond (single bond, double bond, or triple bond) counts as exactly one electron domain.

For example, in $\rm H_2C=C=O$ , the carbon atom at the center of that $\rm C=C=O$ bond has two electron domains:

This carbon atom has two double bonds: one $\rm C=C$ bond and one $\rm C=O$ bond. Even though these are both double bonds, in VSEPR theory, each of them count only as one electron domain.
Keep in mind that there are only four valence electrons in each carbon atom. It can be shown that all four valence electrons of this carbon atom are involved in bonding (two in each of the two double bonds.) Hence, there would be no nonbonding pair around this atom.

In VSEPR theory, electron domains around an atom repel each other. As a result, they would spread out (in three dimensions) as far away from each other as possible. When there are only two electron domains around an atom, the two electron domains would form a straight line- with one domain on each side of the central atom. (To visualize, consider the three atoms in this $\rm C=C=O$ bond as three spheres on a stick. The central $\rm C$ atom would be between the other $\rm C$ atom and the $\rm O$ atom.)

This linear geometry corresponds to a bond angle of $180^\circ$ .

3 0

3 years ago