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Leno4ka [110]
3 years ago
13

1. How many moles of iron is needed to react with sulfur in order to produce 8.6 moles

Chemistry
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

17.2 moles of Iron are required

Explanation:

Based on the chemical reaction:

2Fe + 3S → Fe₂S₃

<em>2 moles of Fe react with 3 moles of S to produce 1 mole of Iron (III) sulfide</em>

<em />

Assuming and excess of sulfur, if we want to obtain 8.6 moles of Fe₂S₃ are required:

8.6 moles Fe₂S₃ * (2 moles Fe / 1mol Fe₂S₃) =

<h3>17.2 moles of Iron are required</h3>
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Which macromolecules are polymers made of nucleotides
VMariaS [17]
Nucleic Acids are polymers made of nucleotideds 
8 0
3 years ago
Read 2 more answers
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

2*n_{acid}=n_{base}

And in terms of concentrations and volumes:

2*M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the volume of acid:

V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL

Best regards.

5 0
3 years ago
What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

K_{a}=1.8\times 10^{-5}

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}

The expression for dissociation constant of acid is:

K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}

1.8\times 10^{-5}=\frac{x^2}{1.8-x}

1.8\left(1.8-x\right)=100000x^2

Solving for x, we get:

<u>x = 0.00568  M</u>

Percentage ionization = \frac{0.00568}{1.8}\times 100=0.32 \%

<u>Option B is correct.</u>

8 0
3 years ago
How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work
strojnjashka [21]
2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
8 0
4 years ago
How to balance equations
dezoksy [38]

start the balancing by writing down how many atoms there are per element. we’ll use this as an example:

C3H8 + O2 --> H2O + CO2

C = 3 C = 1

H = 8 H = 2

O = 2 O = 3

balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:

C3H8 + O2 --> H2O + (3)CO2

now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.

C = 3 C = 3

H = 8 H = 2

O = 2 O = 7

now let’s move on to the hydrogen atoms.

C3H8 + O2 --> H2O + 3CO2

you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.

C3H8 + O2 --> (4)H2O + 3CO2

now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.

C = 3 C = 3

H = 8 H = 8

O = 2 O = 10

now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10

C3H8 + (5)O2 --> 4H2O + 3CO2

the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.

C = 3 C = 3

H = 8 H = 8

O = 10 O = 10

4 0
3 years ago
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