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Troyanec [42]
3 years ago
8

A pharmacist has a 12% solution of boric acid and a 20% solution of boric acid. How much of each must he use to make 80 grams of

a 15% boric acid solution
Chemistry
1 answer:
Iteru [2.4K]3 years ago
5 0

Answer:

  • <u><em>He must use 50g of the 12% solution and 30 g of the 20% solution.</em></u>

<u><em></em></u>

Explanation:

Call x the amount of <em>12% boric acid </em>solution to be used.

  • The content of acid of that is: 0.12x

Since he wants to make <em>80 grams</em> of solution, the amount of <em>20% boric acid</em> solution to be used is 80 - x.

  • The content of acid of that is{ 0.20(80 - x).

The final solution is <em>15% </em>concentrated.

  • The content of boric acid of that is 0.15 × 80 g.

Now you have can write your equation:

         0.12x+0.20(80-x)=0.15\times 80

Solve:

           0.12x + 16 - 0.20x=12\\\\-0.08x=12-16\\\\0.08x=4\\\\x=4/0.08\\\\x=50

That is 50 grams of the 12% solution of boric acid.

Calculate the amount of the 20% solution of boric acid:

         80-x=80-50=30

That is 30 grams.

Then, he must use 50g of the 12% solution and 30 g of the 20% solution.

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MakcuM [25]

Answer : The value of \Delta H_{rxn}  for the reaction is, -135.2 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of ClF_3 will be,

ClF(g)+F_2(g)\rightarrow ClF_3(l)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) 2ClF(g)+O_2(g)\rightarrow Cl_2O(g)+OF_2(g)     \Delta H_1=167.5kJ

(2) 2F_2(g)+O_2(g)\rightarrow 2OF_2(g)    \Delta H_2=-43.5kJ

(3) 2ClF_3(l)+2O_2(g)\rightarrow Cl_2O(g)+3OF_2(g)    \Delta H_3=394.4kJ

We are dividing the reaction 1, 2 and 3 and reversing reaction 3 and then adding all the equations, we get :

(1) ClF(g)+\frac{1}{2}O_2(g)\rightarrow \frac{1}{2}Cl_2O(g)+\frac{1}{2}OF_2(g)     \Delta H_1=\frac{167.5kJ}{2}=83.75kJ

(2) F_2(g)+\frac{1}{2}O_2(g)\rightarrow OF_2(g)    \Delta H_2=\frac{-43.5kJ}{2}=-21.75kJ

(3) \frac{1}{2}Cl_2O(g)+\frac{3}{2}OF_2(g)\rightarrow ClF_3(l)+O_2(g)    \Delta H_3=\frac{-394.4kJ}{2}=-197.2kJ

The expression for enthalpy of formation of ClF_3 will be,

\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(83.75kJ)+(-21.75kJ)+(-197.2kJ)

\Delta H_{rxn}=-135.2kJ

Therefore, the value of \Delta H_{rxn}  for the reaction is, -135.2 kJ

5 0
3 years ago
An impure sample of table salt that weighed 0.8421 g, when dissolved in water and treated with excess AgNO3, formed 2.044 g of A
Lilit [14]

Answer:

99.24%.

Explanation:

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<em>NaCl + AgNO₃ → AgCl(↓) + NaNO₃,</em>

1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.

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no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.

  • Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:

<em>using cross multiplication:</em>

1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.

∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.

  • Now, we can get the mass of puree NaCl in the sample:

mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.

∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.

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3 years ago
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GenaCL600 [577]
 <span>this is a limiting reagent problem. 

first, balance the equation 
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now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops. 

So, the mass of sodium oxide is 

75.5 g</span>
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Answer:

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