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forsale [732]
3 years ago
12

Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Asp

irin, 5 gr." How many mg of aspirin are in a tablet containing 5 gr?
Chemistry
1 answer:
viva [34]3 years ago
8 0

Answer:

325mg of Aspririn

Explanation:

First you should note the information that the problem gives you:

- The bottle of Aspirin has 5gr (grains)

- 1gr(grain) = 65mg (miligrams)

Also, the problem is asking about how many aspirin are in 5 gr (grains), so you should use a conversion factor, as follows:

-First you should put the quantity you need to convert:

5grAspirin

-Then you write the denominator of the conversion factor that must have the same units that you want to convert, in this case gr:

5grAspirin*\frac{}{1grAspirin}

-Then you write the numerator with the units that you want to obtain and the numerical equivalence between the units, in this case:

5grAspirin*\frac{65mgAspririn}{1grAspirin}

-Finally you multiply numerators and divide by denominators:

5grAspirin*\frac{65mgAspririn}{1grAspirin}=325mgAspririn

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The following data was collected when a reaction was performed experimentally in the laboratory.
Lera25 [3.4K]

Answer:

Mass = 279.23 g

Explanation:

Given data:

Number of moles of Fe₂O₃ = 3 mol

Number of moles of Al = 5 mol

Maximum amount of iron produced by reaction = ?

Solution:

Chemical equation:

Fe₂O₃ + 2Al    →     Al₂O₃  +  2Fe

Now we will compare the moles of iron with Al and iron oxide.

                          Fe₂O₃     :       Fe

                             1           :        2

                             3          :       2×3 = 6 mol

                            Al          :          Fe

                              2         :           2

                               5        :           5 mol

The number of moles of iron produced by Al are less so Al is limiting reacting and it will limit the amount of iron so maximum number of iron produced are 5 moles.

Mass of iron:

Mass = number of moles × molar mass

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3 0
3 years ago
Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
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