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bazaltina [42]
3 years ago
9

Which of the following characteristics do Element I and Element I have in common?

Chemistry
1 answer:
Nesterboy [21]3 years ago
5 0

Answer:

Option C. The same number of energy levels.

Explanation:

From the diagram given above, element (i) belong to group 2 while element (ii) belong to group 6.

Also, both element i and ii belong to the same period (i.e period 4). This simply means that both element i and ii have the same number of energy levels.

NOTE: Elements in the same period have the same number of shells of electrons which simply means they have the same energy levels.

You might be interested in
Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hy
Serhud [2]

Answer:

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

Explanation:

The complete reaction when hydroiodic acid and sodium hydrogen carbonate combine, would be as follows -

HI + NaHCO3 ----> NaI + H2O + CO2

net reaction

H2CO3 is highly unstable, and thus decomposes into the water and carbon dioxide you see present as the reactants. If you didn't know already, H2CO3 is also reffered to as carbonic acid. The rest of the elements present on the reactant side are Iodine and Sodium, which is why they are present on the product side as NaI.

Let me include the " physical states " in this reaction as well -

HI ( aq ) + NaHCO3 ( aq ) ----> NaI ( aq ) + H2O ( l ) + CO2 ( g )

Now the complete ionic equation would simply be each compound present as ions in an aqueous solution, so there is no need for an explanation on this step -

H+ ( aq ) + I- ( aq ) + Na+ ( aq ) + HCO3- ( aq ) -------> Na+ ( aq ) + I- ( aq ) + H2O( l ) + CO2 ( g )

The spectator ions in this reaction are I- and Na+, so canceling them out, you would receive the following net ionic equation -

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

<u><em>Hope that helps!</em></u>

4 0
3 years ago
Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

Answer:

M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

Best regards!

5 0
2 years ago
What is the ph of a solution of 1.699 l of 1.25 m hcn, ka = 6.2 x 10-10, and 1.37 moles of nacn?
BlackZzzverrR [31]

The pH of a solution is 9.02.

c(HCN) = 1.25 M; concentration of the cyanide acid

n(NaCN) = 1.37 mol; amount of the salt

V = 1.699 l; volume of the solution

c(NaCN) = 1.37 mol ÷ 1.699 l

c(NaCN) = 0.806 M; concentration of the salt

Ka = 6.2 × 10⁻¹⁰; acid constant

pKa = -logKa

pKa = - log (6.2 × 10⁻¹⁰)

pKa = 9.21

Henderson–Hasselbalch equation for the buffer solution:

pH = pKa + log(cs/ck)

pH = pKa + log(cs/ck)

pH = 9.21 + log (0.806M/1.25M)

pH = 9.21 - 0.19

pH = 9.02; potential of hydrogen

More about buffer: brainly.com/question/4177791

#SPJ4

8 0
11 months ago
What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4
Ivenika [448]

Answer:

130ml of HCl(36%) in 4.90L solution => pH = 1.50

Explanation:

Need 4.90L of HCl(aq) solution with pH = 1.5.

Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺

[HCl(36%)] ≅ 12M in HCl

(M·V)concentrate = (M·V)diluted

12M·V(conc) = 0.032M·4.91L

=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.

Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.

5 0
3 years ago
Hey guys, about to submit my draft for business and chemistry.
drek231 [11]

Answer:

so u can put on hot sauce to spice it up

and give it to ur teacher

8 0
3 years ago
Read 2 more answers
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