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dolphi86 [110]
2 years ago
11

Identify the solute and solvent in a dilute aqueous solution of potassium chloride

Chemistry
1 answer:
erastova [34]2 years ago
6 0
Solute=potassium chloride, solvent=water
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Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g
satela [25.4K]

Answer:

5.6gNa_2SO_4

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4

Best regards.

5 0
3 years ago
Some antacid tables contain aluminum hydroxide. The aluminum hydroxide reacts with stomach acid according to the equation: Al(OH
MatroZZZ [7]

Answer:

26.67 mol HCl

Explanation:

Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O

In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.

To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:

  • 8.89 mol Al(OH)₃ * \frac{3molHCl}{1molAl(OH)_{3}} = 26.67 mol HCl

Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.

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As the gas cools it condenses and becomes a liquid its atoms also become smaller
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amm1812

Answer:

If sunspots are active, more solar flares will result creating an increase in geomagnetic storm activity for Earth.

Explanation:

I ask mom

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