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3 years ago

13

Name the element which has twice the number of electrons in its second shell as in the first shell write it's electronic configu

ration also.

1 answer:

Basile [38]3 years ago

3 0

CARBON! :) it’s the 6th element

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State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA], for each of the fol

melisa1 [442]

When pure HA is added to the buffer, the buffer component ratio and the pH decrease.

<h3>State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA] for the dissolve of pure HA in the buffer.</h3>

When pure HA is added to the buffer, the buffer component ratio and the pH decrease. The added HA increases the concentrations of NA and HA. However, there is a greater relative increase in the concentration of HA. Hence, the ratio of [NaA]/[HA] decreases, causing the solution to become more acidic.

The capacity of a buffer to withstand pH change is measured. The concentration of the buffer's components namely, the acid and its conjugate base determine this ability. Greater buffer capacity is associated with higher buffer concentration.

To learn more about buffer-component, Visit:

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8 0

2 years ago

A cluster of billions of atoms that all have magnetic fields lined up in the same way is known as a

satela [25.4K]

A cluster of billions of atoms that all have magnetic fields lined up in the same way is known as a

<span>The answer is letter C. Magnetic domain.</span>

<span>
The region around a magnet where the magnetic force is exerted is known as its
The answer is letter C. Magnetic field.</span>

6 0

3 years ago

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

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

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution</u>

Formula used :

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

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

<u>(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution</u>

Formula used :

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

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

<u>(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution</u>

Formula used :

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

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

<u>(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution</u>

Formula used :

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

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution</u>

Formula used :

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

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

Match the term with its description. (4 points)

Ne4ueva [31]

Answer:

A. a

B. c

C. b

D. d

Explanation:

3 0

3 years ago

Brainliest to whoever gets it correct.

pochemuha

Answer:

soi

brgm

12ck

3hp akdjskfkoos

8 0

3 years ago