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Likurg_2 [28]
2 years ago
8

Which group 2A element has the largest ionic radius?

Chemistry
2 answers:
zvonat [6]2 years ago
7 0

Explanation:

Group 2A elements are Beryllium, Magnesium, Calcium, Strontium, Barium and Radium.

An atom which is largest in size will have the largest ionic radius. Also, it is known that as we move down a group size of atoms increases.

So, in group 2A Beryllium is smallest in size whereas Radium is the largest in size.

Thus, we can conclude that group 2A element which has the largest ionic radius is radium.

ivann1987 [24]2 years ago
5 0
The answer to the question is radium (Ra)
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How does the idea of a neutron help address the puzzles about the protons in the nucleus and the mass of atoms?
Kisachek [45]

Answer:

hen the number of neutrons is known and the atomic number of an element is known, it becomes easier to determine the approximate mass number by adding the two.

Explanation:

Hope it shelps

6 0
2 years ago
When an iron nail is ground into powder, its mass ____.
Gre4nikov [31]

Explanation:

(a) The mass of an object remains the same always. It is independent of its location. In this case, an iron nail is ground into powder. It means that its mass will remain unchanged. Hence, (a) is the correct option.

(b) Whenever there is a change of color of an object, a chemical reaction occurs. It shows that a reaction happens there. In this problem, a paper turns yellow-brown upon exposure to sunlight. A chemical reaction takes place there. Hence, (b) is the correct option.

3 0
3 years ago
Astudent prepareda calibration curve by plotting absorbance of the standards against the [FeSCN2+] molar concentration (M). The
Nuetrik [128]

Answer:

Explanation:

Chemistry 1B Experiment 7

1-3 5.0 1.5 3.5

Part 2: Determining the equilibrium constant.

Label 5 medium-sized test tubes. Table 7.2 shows the amounts of 2.00 × 10–3

M

Fe(NO3)3 (in 1 M HNO3) solution, 2.00 × 10–3

M KSCN solution, and purified water

that should be added to each tube. Pipet the approximate amount of each solution into

each tube. (Record the exact amount of each solution that you actually add. You will

need to use these actual amounts in your calculations.)

Obtain five separate small pieces of parafilm. Close the top of each test tube with

the parafilm. Mix each solution thoroughly by inverting the test tube several times.

Record your observations.

Measure and record the absorbance of each solution at the 447 nm.

Table 7.2 Composition of solutions for determining the equilibrium constant.

Test Tube

Volume of

2.00 × 10–3

M Fe(NO3)3

in 1 M HNO3 (mL)

Volume of

2.00 × 10–3

M KSCN

(mL)

Volume of

purified water

(mL)

2-1 5.0 1.0 4.0

2-2 5.0 2.0 3.0

2-3 5.0 3.0 2.0

2-4 5.0 4.0 1.0

2-5 5.0 5.0 none

Calculations

Part 1. Graphing the relationship between absorbance and [FeSCN2+].

Assuming that “all” of the SCN–

ions have been converted to FeSCN2+ ions,

calculate [FeSCN2+] in each of the solutions in Part 1. For example, in test tube 1-2, 1.0

mL of a 2.00 × 10–3

M KSCN solution was diluted to 10.0 mL. The concentration of

SCN–

that results from this dilution is the one to use for determining [FeSCN2+].

Because of the 1:1 stoichiometry, that initial concentration of SCN– is equal to

[FeSCN2+].

Plot a full-page graph of the absorbance against the concentration of FeSCN2+ in

all standard solutions. Use a ruler to draw the best straight line that comes closest to each

of your five data points. Your line should pass through (0 M, 0). (Why?) This graph is

your calibration curve. When you measure the absorbance of a solution that contains an

unknown concentration of FeSCN2+ ions, you can use this calibration curve to determine

the unknown concentration

6 0
3 years ago
The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H2SO4 contains more HSO4− ions than SO42−
PIT_PIT [208]

The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H₂SO₄ contains more HSO₄⁻ ions than SO₄²⁻ ions is <u>True.</u> This is best explained by the fact that H₂SO₄ <u>is a diprotic acid where only the first hydrogen completely ionizes.</u>

Why?

H₂SO₄ is a diprotic acid. That means that it has <u>two hydrogen ions</u> to give to the solution. The two dissociation reactions are shown below:

H₂SO₄ + H₂O → HSO₄⁻ + H₃O⁺

HSO₄⁻ + H₂O ⇄ SO₄²⁻ + H₃O⁺

As the arrows show, the first dissociation is complete, meaning that all the sulfuric acid that is present initially is dissociated into HSO₄⁻ and H₃O⁺. However, the second dissociation is incomplete, and it's actually an equilibrium with an acid constant  (Ka)of 1.2×10⁻².

That means that if the initial concentration of H₂SO₄ was 1M, the concentration of HSO₄⁻ is going to be 1M as well, but <u>the concentration of SO₄²⁻ is going to be much less than 1M</u>, according to the dissociation constant.

Have a nice day!

4 0
2 years ago
The standard entropy of liquid methanol at 298K is 126.8 J/K-mol and its heat capacity is 81.6 J/K-mol. Methanol boils at 337K w
USPshnik [31]

Explanation:

First, we will calculate the entropies as follow.

 \Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}} J/K mol

As,   T_{1} = 298 K,         T_{2} = 373.8 K

Putting the given values into the above formula we get,

     \Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}} J/K mol

                 = 81.6 ln (\frac{373.8}{298})

                 = 18.5 J/K mol

Now,

       \Delta S_{3} = \frac{\Delta H_{vap}}{T_{b.p}}

                  = \frac{35270 J}{373.8}

                  = 94.2 J/mol K

Also,

    \Delta S_{4} = C_{p_{4}} ln \frac{T_{2}}{T_{1}}

                = 43.89 \times ln (\frac{800}{373.8})

                = 33.4 J/K mol

Now, we will calculate the entropy of one mole of methanol vapor at 800 K as follows.

   \Delta S_{T} = \Delta S^{o}_{1} + \Delta S_{2} + \Delta S_{3} + \Delta S_{4}

                 = (126.8 + 18.5 + 94.2 + 33.4) J/K mol

                 = 272.9 J/K mol

Thus, we can conclude that the entropy of one mole of methanol vapor at 800 K is 272.9 J/K mol.

7 0
3 years ago
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