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babymother [125]
3 years ago
10

If a liquid has a density of 1200 g/l what is it’s density in charge/ml

Chemistry
1 answer:
Juliette [100K]3 years ago
5 0

Answer:- 1.2\frac{g}{mL}

Solution:- The given density is 1200\frac{g}{L} and it asks to convert it to \frac{g}{mL} .

Here, we could notice that only volume unit is changing ad we know that, 1 L = 1000 mL. Let's make the set up as:

1200\frac{g}{L}(\frac{1L}{1000mL})

= 1.2\frac{g}{mL}

So, the density is 1.2\frac{g}{mL} .

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What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
bulgar [2K]

Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

\Delta T_b = I \times K_b \times (w/M)/ x

\Delta T_b = I \times K_b \times w/Mx

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4 0
3 years ago
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

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