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2 years ago

How many grams of KBr are required to make 550. mL of a 0.115 M KBr solution?

1 answer:

4 0

Molarity is expressed as the number of moles of solute per volume of the solution. For example, we are given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH per 1 L volume of the solution. We calculate as follows:

0.115 M = n mol KBr / .55 L solution

n = 0.06325 mol KBr

mass = 0.06325 mol KBr (119 g / mol) = 7.53 g KBr

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Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord

Marina CMI [18]

Answer:

$M_{acid}=0.563M$

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the molarity of the acid to obtain:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M$

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5 0

2 years ago

A white powder, when heated, produces a colourless gas and a black solid. Is the white powder an element? I need reasons please

NeTakaya

Answer:It is not an element because elements are the purest form of a substance; hence, they are no longer broken down by heating

Explanation:

4 0

2 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago

A buffer is prepared by adding 100 mL of 0.50 M sodium hydroxide to 100 mL of 0.75 M propanoic acid. Is this a buffer solution,

Yanka [14]

The answer: is yes, It is a buffer solution.

first, we need to get moles of sodium hydroxide and propanoic acid:

moles NaOH = molarity * volume

= 0.5M * 0.1 L = 0.05 moles

moles propanoic acid = molarity * volume

= 0.75 M * 0.1 L = 0.075 moles

[NaOH] at equilibrium = 0.05 m

[propanoic acid ] at equilibrium = 0.075 - 0.05 = 0.025 m

when Pka for propanoic acid (given) = 4.89

so by substitution:

∴PH = Pka + ㏒[NaOH]/[propanoic acid ]

∴ PH = 4.89 + ㏒ 0.05 / 0.025

= 5.19

7 0

2 years ago

A scientist wants to create a drug that will kill cancer cells upon absorption in the intestines. In order for the drug to work

Anarel [89]

The testing of the acidities of the compounds to determine the combination that will keep the drug intact before it reaches the intestines is a part of EXPERIMENTATION process. The experimentation is done in order to prove or answer different queries that were raised during the hypothesis making.

8 0

2 years ago

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