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3 years ago

Which of the following elements has the smallest atomic radius

Chemistry

1 answer:

babunello [35]3 years ago

3 0

Answer:

Element with the smallest radius is Carbon.

Explanation:

In a periodic table, atomic radius increases down the group due to addition of a new shell and decreases across the period from left to right due to increasing nuclear charge.

Due to addition of more electrons in same shell and increase of positive charge in nucleus increases attractive forces between electrons and nucleus hence decreasing size.

Carbon and Lithium are present in same period and hence Carbon has smaller size; Potassium and Bromine are present in same period and hence out of the two, Bromine has smaller size.

On comparing Carbon and Bromine, atomic radius increases down the group hence, Carbon has the smallest radius among the four given elements.

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Is my current answer correct? if not, please correct me

MAVERICK [17]

Answer:

Yes you’re correct :)

Explanation:

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3 years ago

Read 2 more answers

Redi placed pieces of meat in several jars. He divided the jars into two groups. He covered the first group of jars with fine cl

OlgaM077 [116]

Answer:

Experiment Question, Independent & Dependent Variable, Null & Alternate Hypothesis

Explanation:

Experiment Question : Whether fine cloth coverage over chicken jars effect flies & their eggs in ambience ?

Independent Variable : The causal variable, ie cloth cover existence on the chicken jar

Dependent Variable : The resultant variable, ie presence of flies & their eggs near jar, meat

Null Hypothesis [H0] : Cloth cover on chicken jar doesn't effects flies' & eggs' presence around jar, chicken.

Alternate Hypothesis [H1] : Cloth cover on chicken jar effects flies' & eggs' presence around jar, chicken.

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3 years ago

The nonvolatile, nonelectrolyte estrogen (estradiol), C18H24O2 (272.40 g/mol), is soluble in chloroform CHCl3.

Artist 52 [7]

Answer: The molarity of solution is 0.274 M and the osmotic pressure of the solution is 6.70 atm

Explanation:

To calculate the molarity of the solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of estrogen = 13.5 g

Molar mass of estrogen = 272.40 g/mol

Volume of solution = 181 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{13.5\times 1000}{272.40\times 181}\\\\\text{Molarity of solution}=0.274M$

Hence, the molarity of solution is 0.274 M

To calculate the osmotic pressure of the solution, we use the equation:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = 0.274 M

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$\pi=1\times 0.274mol/L\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\pi=6.70atm$

Hence, the osmotic pressure of the solution is 6.70 atm

8 0

3 years ago

A student sets up the following equation to solve a problem in solution stoichiometry. (The ? stands for a number the student is

Leni [432]

Answer:

mL * L²

Explanation:

The question in t his problem is to calculate the units of the final answer.

The units in the numerator is mol, ml and L.

The unit in the denominator is mol/L

This leads us to;

Numerator / Denominator = mol * mL * L / (mol / L )

The final units is mL * L²

3 0

3 years ago

A 50.0 mL sample of 0.00200 M AgNO3 is added to 50.0 mL of 0.0100 M NaIO3. What is the equilibrium concentration of Ag in soluti

Nana76 [90]

The equilibrium concentration of Ag⁺ in the solution is : 7.5 * 10⁻⁶ M

Given that :

Ksp for AgIO₃ = 3 * 10⁻⁸

<h3>Determine the equilibrium concentration of Ag in the solution </h3>

First step : Calculate the concentration of Ag⁺ and IO⁻₃ in the solution

[ Ag⁺ ] = ( mmol Ag⁺ / mL solution )

= ( 50 * 0.00200 / 100 mL ) = 0.001 M

[ IO⁻₃ ] = ( mmol IO⁻₃ / mL solution )

= ( 50 * 0.0100 / 100 mL ) = 0.005 M

Next determine the Ionic product ( Q )

Q = [ Ag⁺ ] [ IO⁻₃ ]

= 0.001 * 0.005

= 5 * 10⁻⁶

Since the value of Q is > Ksp a precipitate ( IO⁻₃ ) will be formed after the completion of the precipitation reaction

Therefore the concentration of the excess IO⁻₃ = 0.400 mmol / 100 mL

= 0.004 M

Second step : considering the initial and final concentrations

Initial concentrations ( mol/ L ) Final concentrations ( mol/L )

[ Ag⁺ ] = 0 M [ Ag⁺ ] = x

[ IO⁻₃ ] = 0.004 M [ IO⁻₃ ] = 0.004 + x

Final step : Determine the equilibrium concentration of Ag in the solution

Ksp = 3 * 10⁻⁸ = [ Ag⁺ ] [ IO⁻₃ ]

= ( x ) ( 0.004 + x )

Therefore x = 7.5 * 10⁻⁶ ( Equilibrium concentration of Ag in the solution )

Hence we can conclude that The equilibrium concentration of Ag⁺ in the solution is : 7.5 * 10⁻⁶ M

Learn more about equilibrium concentration : brainly.com/question/13414142

6 0

2 years ago