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Marat540 [252]
3 years ago
11

Which element in the third period would you expect to have the larger atomic radius, sodium (Na) or Sulfur (S)?

Chemistry
2 answers:
jeyben [28]3 years ago
7 0

<u>Answer: </u>The correct answer is Option B.

<u>Explanation:</u>

The given elements that is sodium and sulfur are the elements which belong to the same period that is third period.

As, the electrons are added up to the same shell across a period, the number of protons also increase and hence, they attract the electrons more effectively. This means that the effective nuclear charge also increases and the size of the sulfur thus shrink and hence, the size (atomic radius) of sodium is the largest amongst third period.

Therefore, the correct answer is Option B.

Doss [256]3 years ago
3 0

Answer:

B

Explanation:

As, the electrons are added up to the same shell across a period, the number of protons also increase and hence, they attract the electrons more effectively. This means that the effective nuclear charge also increases and the size of the sulfur thus shrink and hence, the size (atomic radius) of sodium is the largest amongst third period.

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How do I find the formula for Phosphorus Tetroxide
salantis [7]
Formula= P4O6
You are going to want to flip the elements subscripts with one another.
P6O4
P will just stay as phosphorus, but since oxygen has 4 atoms, itbwill become tetroxide
7 0
3 years ago
Suppose that 0.50 grams of barium-131 are administered orally to a patient. Approximately how many milligrams of the barium woul
nalin [4]

Two months later 13.8 milligrams of the barium-131 still be radioactive.

<h3>How is the decay rate of a radioactive substance expressed ? </h3>

It is expressed as:

A = A_{0} \times (\frac{1}{2})^{t/T}

where,

A = Amount remaining

A₀ = Initial Amount

t = time

T = Half life

Here

A₀ = 0.50g

t  = 2 months = 60 days

T = 11.6 days  

Now put the values in above expression we get

A = A_{0} \times (\frac{1}{2})^{t/T}

   = 0.50 \times (\frac{1}{2})^{60/11.6}

   = 0.50 \times (\frac{1}{2})^{5.17}

   = 0.50 × 0.0277

   = 0.0138 g

   = 13.8 mg          [1 mg = 1000 g]

Thus from the above conclusion we can say that Two months later 13.8 milligrams of the barium-131 still be radioactive.

Learn more about the Radioactive here: brainly.com/question/2320811

#SPJ1

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Suppose that 0.50 grams of ban that 0.50 grams of barium-131 are administered orally to a patient. Approximately many milligrams of the barium would still be radioactive two months later? The half-life of barium-131 is 11.6 days.

3 0
1 year ago
Ill give the brainliest answer to whoever helps me with this equation
vampirchik [111]

Answer: The percent yield for the NaBr is, 86.7 %

Explanation : Given,

Moles of FeBr_3 = 2.36 mol

Moles of NaBr = 6.14 mol

First we have to calculate the moles of NaBr

The balanced chemical equation is:

2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr

From the reaction, we conclude that

As, 2 moles of FeBr_3 react to give 6 moles of NaBr

So, 2.36 moles of FeBr_3 react to give \frac{6}{2}\times 2.36=7.08 mole of NaBr

Now we have to calculate the percent yield for the NaBr.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%

Therefore, the percent yield for the NaBr is, 86.7 %

6 0
3 years ago
What type of charge does a neutron have
REY [17]
It is electric force
8 0
3 years ago
What is the percentage by mass of nacl in a saturated solution of sodium chloride at 50 c?
uranmaximum [27]
In this solution we are having two components i.e. NaCl and H₂O. So the %age mass of NaCl is calculated by following formula,

%age mass of NaCl  =  (Mass of NaCl / Mass of NaCl + Mass of H₂O) × 100     ------ (1)

Calculating Mass of NaCl at 50°C;

Solubility of NaCl was searched online and was found 36.69 g / 100 mL of water at 50 °C.

Calculating Mass of 100 mL H₂O at 50°C;

Density of H₂O at 50 °C is 0.988 g/ml, so for 100 mL
As,
                           Density  = Mass / Volume

                           Mass  =  Density × Volume

                           Mass  =  0.988 g/mL × 100 mL

                           Mass  =  98.8 g

Putting Masses of NaCl and H₂O in eq. 1,

              %age mass of NaCl  =  (36.69 g / 36.69 g + 98.8 g) × 100

              %age mass of NaCl  =  (36.69 g / 135.49 g) × 100

              %age mass of NaCl  =  27.07 %
4 0
3 years ago
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