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serg [7]
3 years ago
15

Which represents the self-ionization of water at 25°C? H2O + H2O 2H2 + O2 H2O + H2O H2O2 + H2 H2O + H2O 4H+ + 2O2- H2O + H2O H3O

+ + OH-
Chemistry
2 answers:
artcher [175]3 years ago
8 0

Answer: Option (d) is the correct answer.

Explanation:

When molecules of same substance combine together to result in the formation of ions then this type of reaction is known as self-ionization.

For example, H_{2}O + H_{2}O \rightarrow H_{3}O^{+} + OH^{-}

So, here two molecules of water are combining and they result in the formation of hydronium ion (H_{3}O^{+}) and hydroxide ion (OH^{-}).

Hence, this reaction is a self-ionization reaction.

Vitek1552 [10]3 years ago
6 0
The fourth answer is correct 
H20 + H20  H30 + 0H-
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You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
An unknown noble gas has a density of 5.84 g dm-3 at STP. Calculate its molar mass, and so identify the gas.
bagirrra123 [75]

The noble gas is Xenon and its molar mass is 131 g/mol.

<h3>What is the molar mass of the noble gas?</h3>

The molar mass of the noble gas is determined as follows;

Let molar mass of unknown gas be M, and mass of gas be m

Density of the noble gas, ρ = 5.8 g/dm³

density = m/V

At STP;

  • temperature, T = 273.15 K
  • pressure, P = 1 atm
  • molar gas constant, R = 0.0821 L.atmK⁻¹mol⁻¹

From ideal gas equation:

PV = nRT

where n = m/M

PV = mRT/M

M = mRT/PV

M =  0.0821 * 273.15 * 5.84/1

Molar mass of the noble gas = 131 g/mol

The noble gas is Xenon which has molar mass approximately equal to 131 g/mol.

Learn more about molar mass at: brainly.com/question/837939

#SPJ1

6 0
1 year ago
what has 4 valence electrons, shows properties of metals and non metals, and has 32 protons in the nucleus of each atom
12345 [234]
Here is your answer

\bold{Germanium}

REASON:

Elements which have 4 valence electrons are generally metalloids.

The metalloids show the properties of both metals and non-metals.

We know that,

no. of protons= Atomic number

So,

Atomic no.= 32

Hence,

The element is Germanium which is a metalloid with 4 valence electrons and has 32 protons in nucleus of each atom because it has atomic no. 32

HOPE IT IS USEFUL
6 0
3 years ago
1. If the pressure of a fixed volume of gas decreased in a sealed container, what variable would you think changed? did this var
grandymaker [24]

Answer:

Temp decreased

Explanation:

If the container is sealed and the pressure DEcreases, then the temperature DEcreased

PV = n RT     n  R   V   are constant   if  P goes down then so does T    

8 0
1 year ago
Write a balanced equation for the complete combustion of each compound. a. formaldehyde (CH2O(g)) b. heptane (C7H17(l)) c. benze
Yuri [45]

Answer:

a. formaldehyde (CH₂O(g)): CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

b. heptane (C₇H₁₇(l)): 4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l))​: C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

Explanation:

In a reaction of combustion, a hydrocarbon compound (composed of C, H and O) reacts with oxygen gas (O₂). The <u>complete</u> combustion of a hydrocarbon - such as formaldehyde, heptane, and benzene - produces carbon dioxide (CO₂) and water (H₂O).  

Thus, we write the reactants and products for each combustion reaction and then we balance the atoms: C, H, and O.

a. formaldehyde (CH₂O(g)):

CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

The chemical equation is already balanced with the coefficients 1: we have the same number of C atoms (1), H (2) and O (3) on both sides of the equation.

b. heptane (C₇H₁₇(l)):

C₇H₁₇(l) + O₂(g) → CO₂(g) + H₂O(g)

Here we have to write a coefficient 28 for CO₂ to balance the C atoms in the products side, and for C₇H₁₇ we write 28/4 = 7. With similar reasoning we found the coefficients for O₂ and H₂O:

4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l))​:

C₆H₆(l) +  O₂(g) → CO₂(g) +  H₂O(g)

First, we write a coefficient of 6 for CO₂ to balance the C atoms. Then, we have to balance H atoms: we write a coefficient 3 in H₂O. Now, we have 12 + 3 = 15 atoms of O on the reactants side. So, we write a half of these number of atoms in the coefficient for O₂: 15/2. We obtain the balanced equation:

C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

5 0
2 years ago
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