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Sphinxa [80]
3 years ago
6

The electron configurations of two unknown elements X and Y are shown . X: 1s^ 2 2s^ 2 2p^ 6 3s^ 1 Y: 1s^ 2 2s^ 1 Which statemen

t is most likely correct about the two elements ?
Chemistry
2 answers:
PIT_PIT [208]3 years ago
8 0

Answer:

See below.

Explanation:

They both have 1 valency electron so will be metallic and in the same Group (Group 1) of the Periodic Table, so will have similar properties.

lozanna [386]3 years ago
7 0

Answer:

This question is incomplete

Explanation:

  • X has an atomic number of 11 and hence it's sodium (Na) while Y has an atomic number of 3 and hence it is Lithium (Li).
  • The atomic mass of X is 23 while that of Y is 7
  • X has three electron shells while Y has two electron shells
  • X and Y cannot form a compound as there ions are both positively charged (cations). However, there ions both have a charge of +1; hence are univalent positive metals.
  • They are both metals and are found in group 1 of the periodic table
  • The are highly reactive metals because they have just one electron in there outermost shell
  • They react with halogens in group 7 to form salts
  • They dissolve in water to form alkaline

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A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added
Ulleksa [173]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.306M

[C_6H_5NH_2]=0.418M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

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

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

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

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

8 0
3 years ago
A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo
Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 15.5 mmHg

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

Mass of solute (insulin) = 33 mg = 0.033 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = 62.364\text{ L.mmHg }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0
3 years ago
g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0
2 years ago
What is the role of a consumer in the flow of energy through a food chain?
pogonyaev

Answer:

The role of consumers in an ecosystem is to obtain energy by feeding on other organisms and sometimes transfer energy to other consumers. Changes that affect consumers can impact other organisms within the ecosystem.

Explanation:

Hope this helped Mark BRAINLIEST!!!!

4 0
3 years ago
What type of cells does the human reproductive system contain?
yuradex [85]
Blood cells, glucose , skin cells and etc .
7 0
3 years ago
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