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4 years ago

When water is decomposed into hydrogen and oxygen, these gases cannot combine again to reform water?

1 answer:

6 0

This statement is false. When water is decomposed into hydrogen and oxygen; these gases are in a position of combining again to reform water. Water is a molecule consisting of one atom of oxygen bound to two hydrogen atoms. The atoms are joined by covalent bonds; however they have addition hydrogen bonds between the partial negative oxygen and the partial positive hydrogen atom.

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As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the po

Ganezh [65]

Answer:

pH = 2.03

Explanation:

The pH can be calculated using the following equation:

$pH = -log [H_{3}O^{+}]$ (1)

The concentration of H₃O⁺ is calculated using the dissociation constant of the next reaction:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺

1.00 M

$K_{a} = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$

Solving the above equation for H₃O⁺, we have:

$[H_{3}O^{+}] = \frac{Ka*[CH_{3}COOH]}{[CH_{3}COO^{-}]}$ (2)

The dissociation constant is equal to:

$pKa = -log(Ka) \rightarrow Ka = 10^{-pKa} = 10^{-4.76} = 1.74 \cdot 10^{-5}$

Now, by solving the equation of the solubility product for Herbigon, we can find [CH₃COO⁻]:

CH₃COOX ⇄ CH₃COO⁻ + X⁺

5.00x10⁻³ M

$K_{sp} = [CH_{3}COO^{-}][X^{+}]$

$[CH_{3}COO^{-}] = \frac{K_{sp}}{[X^{+}]} = \frac{9.40 \cdot 10^{-6}}{5.00 \cdot 10^{-3}} = 1.88 \cdot 10^{-3} M$

By entering the values of [CH₃COO⁻] and Ka, into equation (2) we can calculate [H₃O⁺]:

$[H_{3}O^{+}] = \frac{1.74 \cdot 10^{-5}*[1.00]}{[1.88 \cdot 10^{-3}]} = 9.26 \cdot 10^{-3} M$

Hence, the pH is:

$pH = -log [H_{3}O^{+}] = -log [9.26 \cdot 10^{-3}] = 2.03$

Therefore, the pH must be 2.03 to yield a solution in which the concentration of X⁺ is 5.00x10⁻³M.

I hope it helps you!

6 0

4 years ago

How can you determine the specific heat of a metal using a calorimeter? make a hypothesis

Sever21 [200]

Answer:

One can determine the specific heat of the metal through using the clarimeter, water, thermometer and using heat equations.

Explanation:

You can learn about heat effects and calorimetery through a simple experiment by boiling water and heating up the metal in it. Then, pour it into your calorimeter and the heat will flow from the metal to the water. The two equlibria will meet: the metal will loose heat into its surroundings (the water) and teh water will absorb the heat. The heat flow for the water is the same as it is for the metal, the only difference being is the negative sign indicating the loss of the heat of the metal.

In terms of theromdynamics, we can deteremine the heat flow for the metal becasue it would be equal to the mangnitued but opposite in direction. Thus, we can say that the specific heat of water qH2O = -qmetal.

4 0

3 years ago

At what temperature would a 1.30 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86

Setler [38]

Answer:

-4.59°C

Explanation:

Let's see the formula for freezing point depression.

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Freezing constant. For water if 1.86°C/m

m = molality (moles of solute in 1kg of solvent)

i = Van't Hoff factor.

0°C - Freezing T° of solution = 1.86°C /m . 1.30m . 1.9

Freezing T° of solution = - (1.86°C /m . 1.30m . 1.9)

Freezing T° of solution = - (1.86°C/m . 1.30m . 1.9) → -4.59°C

NaCl → Na⁺ + Cl⁻

4 0

3 years ago

How much heat is involved in condensing 15.0 g of ethanol at its boiling point, then cooling it from its boiling point to a fina

Mekhanik [1.2K]

Answer:

Explanation:

3 0

3 years ago

To what final concentration of NH3 must a solution be adjusted to just dissolve 0.060 mol of NiC2O4 (Ksp = 4×10−10) in 1.0 L of

ololo11 [35]

Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- 
NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp 
Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf 

Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48

K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶
0.48 = (0.060)² / [NH3]⁶ ... (dissolved C2O4 2- = 0.060M)
[NH3]⁶ = (0.060)² / 0.48 = 0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075

NH3 = 0.44 M

3 0

3 years ago