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Fittoniya [83]
2 years ago
10

Identify the reactant that is a Brønsted−Lowry acid in the following reaction: HI(aq)+H2O(l)→I−(aq)+H3O+(aq) Express your answer

as a chemical formula. nothing Request Answer Part B Identify the reactant that is a Brønsted−Lowry base in the following reaction: HI(aq)+H2O(l)→I−(aq)+H3O+(aq) Express your answer as a chemical formula. nothing Request Answer Part C Identify the reactant that is a Brønsted−Lowry acid in the following reaction: F−(aq)+H2O(l)⇌HF(aq)+OH−(aq) Express your answer as a chemical formula. nothing Request Answer Part D Identify the reactant that is a Brønsted−Lowry base in the following reaction: F−(aq)+H2O(l)⇌HF(aq)+OH−(aq) Express your answer as a chemical formula.
Chemistry
1 answer:
Fed [463]2 years ago
6 0

Answer:

(1) HI is Bronsted-Lowry acid and H_{2}O is Bronsted-Lowry base.

(2) H_{2}O is Bronsted-Lowry acid and F^{-} is Bronsted-Lowry base.

Explanation:

A Bronsted-Lowry acid is a species which donates proton (H^{+}).

A Bronsted-Lowry base is a species which accepts proton (H^{+}).

(1) HI(aq)+H_{2}O(l)\rightarrow I^{-}(aq)+H_{3}O^{+}(aq)

Reactants: HI and H_{2}O

Here HI donates proton and H_{2}O accepts proton.

Hence HI is Bronsted-Lowry acid and H_{2}O is Bronsted-Lowry base.

(2) F^{-}(aq)+H_{2}O(l)\rightleftharpoons HF(aq)+OH^{-}(aq)

Reactants: F^{-} and H_{2}O

Here H_{2}O donates proton and F^{-} accepts proton.

Hence H_{2}O is Bronsted-Lowry acid and F^{-} is Bronsted-Lowry base.

You might be interested in
Assume the following substances are soluble in water.
solmaris [256]

Ammonium carbonate will form 3 moles of ions.

Methyl alcohol will form 0 moles of ions.

Methane will form 0 moles of ions.

Aluminum sulfite will form 3 moles of ions.

Hydrobromic acid will form 2 moles of ions.

Explanation:

One mole of ammonium carbonate will form 3 moles of ions when dissolved in water.

(NH₄)₂CO₃ (s) + H₂O (l) → 2 NH₄⁺ (aq) + CO₃²⁻ (aq) + H₂O (l)

One mole of methyl alcohol will form 0 moles of ions when dissolved in water.

(actually it form ions because of its acidic behavior but they are at the order of 10⁻⁷ moles, however in the framework of this question we may say that there are none)

One mole of methane will form 0 moles of ions when dissolved in water.

Methane does not react with water (in normal conditions) so will not form ions.

One mole of aluminum sulfite will form 3 moles of ions when dissolved in water.

Al₂SO₃ (s) + H₂O (l) → 2 Al₃⁺ (aq) + SO₃²⁻ (aq) + H₂O (l)

One mole of hydrobromic acid will form 2 moles of ions when dissolved in water.

HBr (l) +  H₂O (l) → Br⁻ (aq) + H₃O⁺ (aq)

Learn more about:

solvation of ions

brainly.com/question/5384053

#learnwithBrainly

8 0
2 years ago
Movement of the ___<br> creates the London dispersion forces.
Tanzania [10]

Answer: electrons

Explanation: moving electrons cause momentarily charge

Distribution on molecule. This distribution induces similar distribution to

Adjacent molecule.

7 0
2 years ago
3. According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density o
velikii [3]

Answer:

a) 11.64 M

b) 43 mL

c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

m = 84*20.37 = 1,711.08 g

m = 1.7 kg

6 0
2 years ago
Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0
2 years ago
Read 2 more answers
The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution i
bija089 [108]

Answer:

110ml

Explanation:

<em>Using the dilution equation, C1V1 = C2V2</em>

<em>Where C1 is the initial concentration of solution</em>

<em>C2 is final concentration of solution</em>

<em>V1 is intital volume of solution</em>

<em>V2 is final volume of solution.</em>

From the question , C1=6M, C2=0.5M, V1=10ml, V2=?

V2 =\frac{C1V1}{C2}

V2 =\frac{10*6}{0.5}

V2 =120ml

volume of water added = final volume -initial volume

                                    = 120-10

                                   =110ml

3 0
2 years ago
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