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3 years ago

How many moles are in 17.2g k2s

Chemistry

1 answer:

Natalija [7]3 years ago

3 0

Answer: 0.156 mol

Explanation:

To find the moles of 17.2 g K₂S, we need to know the molar mass to convert.

$17.2g*\frac{mol}{110.256 g} =0.156 mol$

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How many elements does a compound need to form?

Vinvika [58]

Answer:

two

Explanation:

When two distinct elements are chemically combined, chemical bonds form between their atoms. the result is called a chemical compound

4 0

2 years ago

Read 2 more answers

Could someone help me with all of these questions. Also plz if you don't know the answer plz don't put something that isn't righ

densk [106]

Q3.) acidic
q4.)that the solution is acidic
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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 $4.58\times 10^{-8}M$

5 0

3 years ago

A gas mixture contains oxygen, nitrogen, and carbon dioxide. It has a pressure of 250 mmHg. If the pressure of oxygen is 50 mmHg

kondor19780726 [428]

Answer:

$P_{CO_2}=25mmHg$

Explanation:

Hello there!

In this case, according to the Dalton's law, which explains that the total pressure of a gaseous system equals the sum of the partial pressures of the gases composing, for the gaseous mixture composed by oxygen, nitrogen and carbon dioxide it would be possible to write:

$P_{tot}=P_{N_2}+P_{O_2}+P_{CO_2}$

Now, given the pressure of the system and those of oxygen and nitrogen, we calculate that of carbon dioxide as shown below:

$P_{CO_2}=P_{tot}-P_{N_2}-P_{O_2}\\\\P_{CO_2}=250mmHg-50mmHg-175mmHg\\\\P_{CO_2}=25mmHg$

Best regards!

4 0

2 years ago

What is the volume (in cubic inches) of 3.6 lb of titanium? The density of titanium is 4.51 g/cm3

SashulF [63]

Answer: 362,07 cm3
To answer this question you need to convert the lb into gram first. One lb equal to 453.592g, so: 3.6lb x 453.592gram/lb= 1632.9312gram.
Now we have mass(1632.9312g), density (4.51g/cm3). Volume is mass divided by density. The equation would be:

Volume= mass/density
Volume = 1632.9312gram / (4.51g/cm3)= 362,07 cm3

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2 years ago