20 points!! Please help ASAP!

How many grams of KBr are required to make 350. mL of a 0.115 M KBr solution?

Vsevolod [243]

0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 × 0.35) = 04025 moles
From the formula of molarity moles = molarity × volume in liters
1 mole of KBr is equivalent to 119 g
Therefore, the mass = 0.04025 × 119 g = 4.79 g

6 0

3 years ago

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A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago

How would you prepare a 1.00L solution of 2.00 M sodium hydroxide<br> (NaOH)*

Ede4ka [16]

Answer:

Explanation:

Molar mass of sodium hydroxide is = 100

any compund with its molar mass dissovled in 1L itres solution gives rise to 1M solution

40 g ------> 1L ---->1M

Xg ------>1L------>2M

X g= 40*1*2/1*1 =80 gram

3 0

3 years ago

Which of these elements exist as a liquid at normal Earth temperatures? Select all that apply. bromine sodium oxygen mercury

Anni [7]

Mercury exist as liquid at normal earth tempertures

4 0

2 years ago

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Convert 400g of phosphoric acid to moles.

Katarina [22]

Answer: 4.08183 or you can round it to 4

Explanation:

3 0

3 years ago