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Annette [7]
3 years ago
8

A cough syrup contains 5.0% ethyl alcohol, c2h5oh, by mass. if the density of the solution is 0.9928 g/ml, determine the molarit

y of the alcohol in the cough syrup
Chemistry
1 answer:
WARRIOR [948]3 years ago
5 0
To answer the question above, let us a basis of the 1000 mL or 1 L. 
                 volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%. 
               mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol). 
                       n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
                       molarity = 1.08 mol/ 1 L = 1.08 M
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7 0
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Which best describes an oxidizing agent?
stellarik [79]

Answer:B) a reactant that undergoes reduction

Explanation:

Oxidation reaction is defined as the reaction in which a substance looses electrons. The oxidation state of the substance increases during oxidation.

M\rightarrow M^++e^-

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced during reduction.

N^++e^-\rightarrow N

Overall reaction: M+N^+\rightarrow M^++N

The substance M which itself gets oxidized, reduces other and is called as reducing agent. The substance N which itself gets reduced, oxidizes other and is called as oxidizing agent.

6 0
3 years ago
Read 2 more answers
Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.
Ronch [10]

Answer : The limiting reagent is O_2

Solution : Given,

Moles of methane = 2.8 moles

Moles of O_2 = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 5 moles of O_2 react with \frac{5}{2}=2.5 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is O_2

8 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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luda_lava [24]

Answer:

D.) 4

Explanation:

Al + 3 O₂ --> 2 Al₂O₃

In the equation, there are:

<em>Reactants:</em> 1 Al and 6 O

<em>Products:</em> 4 Al and 6 O

To determine how many atoms there are of each, you multiply the coefficients by the subscripts attached to the atoms. While the question doesn't ask, as you can see, the oxygen atoms are balanced because there is an equal amount on both sides. To balance the aluminum atoms, you can get 4 aluminum atoms on the reactants side by using a coefficient of 4.

The new equation would look this this:

4 Al + 3 O₂ --> 2 Al₂O₃

There are now:

<em>Reactants:</em> 4 Al and 6 O

<em>Products:</em> 4 Al and 6 O

4 0
2 years ago
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