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2 years ago

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Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod

ine solution

1 answer:

son4ous [18]2 years ago

8 0

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

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A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago

A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete

lys-0071 [83]

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

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2 years ago

A soft drink contains 11.5% sucrose (C12H22O11) by mass. How much sucrose, in grams, is contained in 355 mL (12 oz) of the soft

luda_lava [24]

Answer:

42.5 g

Explanation:

Calculate the mass of the soft drink given the density and volume:

355 mL × 1.04 g/mL = 369.2 g

Now calculate the mass of sucrose given the percentage:

0.115 × 369.2 g = 42.46 g

Rounded to 3 significant figures, the mass is 42.5 g.

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2 years ago

Read 2 more answers

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 ;

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

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

$[H^{+} ] =0.0014$

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

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

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2 years ago

Which is stronger - the attractive forces between water molecules and chromium and chloride ions, or the combined ionic bond str

Allisa [31]

Out of the two, the forces between water molecules and chromium and chloride ions is greater. This is proven by the fact that chromium chloride is slightly soluble in water, about 565 grams per liter.
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2 years ago