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3 years ago

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Litmus paper can be used to determine the _______ of a substance. A. relative volume B. relative pH C. temperature D. chemical f

ormula

1 answer:

fgiga [73]3 years ago

5 0

Answer:

B

Explanation:

Litmus paper determines the relative pH

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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

What features make a question scientific, and why are some questions beyond the boundaries of scientific investigation?

jeka57 [31]

The scientific method

8 0

3 years ago

Identify the Lewis acid and Lewis base from among the reactants in each of the following equations. Match the words in the left

goldfiish [28.3K]

Answer:

1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻

2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻

3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃

Hope this helps.

8 0

3 years ago

Help needed Assignment is due Justify that H2SO4 is Arrhenius acid and KOH is Arrhenius base.

irakobra [83]

Answer:

<u><em>Arrhenius Acid:</em></u>

According to Arrhenius concept, Acids are proton donors.

Since H₂SO₄ have a proton (H⁺ ion) and it can donate it to be made a sulphate ion, So it is an Arrhenius acid.

See the following reaction =>

<u><em>H₂SO₄ + H₂O => HSO₄ + H₃O⁺</em></u>

<u><em>Arrhenius Base:</em></u>

An Arrhenius base is a a proton acceptor.

KOH accepts the proton to to made to KOH₂ and a proton acceptor.

See the following reaction =>

<u><em>KOH + H₂o => KOH₂ + OH⁻</em></u>

<u><em></em></u>

6 0

3 years ago

If you have 250 mL of 0.75 M LINO2.<br> How many grams of this substance<br> would you have?

Vesna [10]

Mass of this substance = 9.928 g

<h3>Further explanation</h3>

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So to find the number of moles can be expressed as

$\boxed{\bold{n=V\times M}}$

mol of substance -Lithium nitrite - LiNO₂ :

V = 250 ml = 0.25 L

M = 0.75 M

$\tt n=0.25\times 0.75=0.1875$

So mass of this substance - LiNO₂ (MW=52,947 g/mol) :

$\tt mass=mol\times MW\\\\mass=0.1875\times 52,947 g/mol\\\\mass=9.928~g$

4 0

3 years ago