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den301095 [7]
3 years ago
9

Which condition applies when water boils at 100°c? air pressure equals a specific value (101.3 kpa) air pressure is less than it

s value at the triple point of water air pressure is zero, as in a vacuum vapor pressure is zero
Chemistry
2 answers:
Alinara [238K]3 years ago
8 0
Water normally boils when the vapor pressure of water is equal to the external pressure. That is, at 100 degree Celsius, the external vapor will be 101.3 kpa. The correct answer is A.
FrozenT [24]3 years ago
4 0

The correct answer is air pressure equals a specific value(101.3 kpa)

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The plants shown bloom in summer. Which statement explains why the plants bloom in summer? The nights are shorter. The nights ar
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Which statement best describes the role that gravity played in the formation of our solar system?
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An atom X has three electrons in its outermost shell. Which ion will most likely be formed by X?
DedPeter [7]

Answer: X3+

Explanation:

Every atom aim to achieve stability by receiving electrons or giving their valence electrons in order to have a complete outermost shell of 2 (duplet) or 8 (octet structure).

In this case, the atom X will easily give off its three valence electrons to another atom(s), thereby forming a trivalent positive ion (X3+) with a stable duplet or octet structure (i.e an outermost shell with 2 or 8 electrons).

X --> X3+ + 3e-

Thus, due to the give away of three electrons (3e-), the atom X becomes X3+.

6 0
3 years ago
At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration
wlad13 [49]

Answer:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.03901 mol/liter
  • [Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

  • Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
  • Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

  • Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
  • Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

            H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

I            0.0481      0.0326            0

C              -x                 -x              +x

E          0.0481-x    0.0326-x         x

<u />

<u>4. Equilibrium expression</u>

       K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}

     0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}

<u />

<u>5. Solve:</u>

            x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0

Use the quadatic formula:

x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}

The positive result is x = 0.00909

Thus the concentrations are:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
  • [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0
3 years ago
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