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2 years ago

7

As the first five elements in Group 14 are considered in order from top to bottom, there are changes in both the

1 answer:

aleksandrvk [35]2 years ago

4 0

Answer:

A

Explanation:

beacause in both the electronegativity values and number of first she'll electrons

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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

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3 years ago

Calculate the molar mass of the acid.

JulijaS [17]

Citric acid has the molecular formula C6H8O7 so you can add the molar masses of the elements from the periodic table. C has a molar mass of 12.01 g/mol, H has 1.01 g/mol and O has 15.999 g/mol. Now you calculate the total molar mass= (6*12.01 + 8*1.01 + 7*15.999). This yields a molar weight of 192.124 g/mol (anhydrous)

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3 years ago

The chlorination of the iron (III) ion is an endothermic reaction.

snow_tiger [21]

Answer:

d

Explanation:

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2 years ago

Resistance of a material being scratched in known as

Nastasia [14]

Friction is the answer

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2 years ago

You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.

Nitella [24]

The question requires us to use the dilution formula $M_iV_i=M_fV_f$ , where $M_i$ and $V_i$ are the stock concentration and volume respectively, then $M_f$ and $V_f$ are the dilute concentration and volume respectively.

a. $C_s_t_o_c_k$ = 1.5 M KCI, $C_d_i_l_u_t_e$ =0.20M KCl, $V_d_i_l_u_t_e$ =250ml KCl

b. $M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml$

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.

8 0

3 years ago