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4 years ago

What is the concentration of a solution which contains 0.446 g of C6H14N4O2 IN 1.250 L of solution? M

Chemistry

1 answer:

dezoksy [38]4 years ago

4 0

Answer:

0.0021 M

Explanation:

Given data:

Mass of C₆H₁₄N₄O₂ = 0.446 g

Volume of solution = 1.250 L

Concentration/Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will calculate the number of moles of solute.

Number of moles = mass/molar mass

Number of moles = 0.446 g/ 174 g/mol

Number of moles = 0.0026 mol

Molarity:

Molarity = 0.0026 mol / 1.250 L

Molarity = 0.0021 M

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Study the reaction. X(s) + 2B(aq)→ X+(aq) + B2(g) Under which circumstances will the reaction rate be highest? A) the solid X is

Tems11 [23]

Answer:

Option (A) the solid X is ground to a fine powder.

Explanation:

X(s) + 2B(aq) → X+(aq) + B2(g)

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4 years ago

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Answer:

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3 years ago

Is volume and mass size-dependent

Otrada [13]

A size dependent property is a physical property that changes when the size of an object changes.

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3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

Which statement correctly describes the actual yield and the theoretical yield of a reaction? (1 point)

Furkat [3]

Answer:

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Explanation:

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8 0

3 years ago