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Gwar [14]
3 years ago
6

Write a balanced molecular equation for hydrochloride acid solution combined with a potassium hydroxide solution, an acid base r

eaction occurs
Chemistry
1 answer:
qaws [65]3 years ago
7 0

Answer:

HCl (aq) + KOH (aq) --------> KCl (aq) + H2O (l)   [balanced molecular equation]

Explanation:

Balanced molecular equation has all the components of reaction written as "molecules" thus the name molecular (the other equations break appropriate components into ions, those are total ionic or net ionic equation)

We know hydrochloride acid solution is written as HCl (aq) and potassium hydroxide solution is written as KOH (aq). We know from our knowledge that when acids react with bases they make salt and water, so now we react!

HCl (aq) + KOH (aq) --------> KCl (aq) + H2O (l)   [balanced molecular equation]

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Fantom [35]

Answer:

NaHCO₃

Explanation:

NaHCO₃ is the chemical formula for sodium bicarbonate or baking soda. It has one sodium atom, one hydrogen atom, one carbon atom, and three oxygen atoms.

Have a good day and stay safe!

6 0
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How are acids and alkalis<br> similar?
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Answer:h

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The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com
Elanso [62]

Answer:

(a) Ksp=4.50x10^{-7}

(b) Ksp=1.55x10^{-6}

(c) Ksp=2.27x10^{-12}

(d) Ksp=1.05x10^{-22}

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)

Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L}   )^2\\\\Ksp=4.50x10^{-7}

(B) Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)

Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}

(C) NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)

Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}

(D) La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)

Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}

Best regards.

7 0
4 years ago
Calculate how many moles are in 63.1 grams FeCl3.
Lynna [10]

Answer: 1 grams FeCl3 = 0.0061650760770388 mole using the molecular weight calculator and the molar mass of FeCl3.

Explanation:

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3 years ago
How many atoms of Hydrogen H are in 2 molecules of Octane 2C8H18
antiseptic1488 [7]

Answer:

16

Explanation:

2x8

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3 0
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