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Vlad1618 [11]
3 years ago
14

For the following list of acids, rank the acids in strength from weakest acid to strongest acid.

Chemistry
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

CH3OH < FCH2OH < F2CHOH < F3COH

Explanation:

Let us recall that, for a carboxylic acid, the dissociation of the acid yields;

RCOOH ⇄RCOO^- + H^+

The ease of dissociation and release of the hydrogen ion depends on the nature of the group designated R.

When R is is a highly electronegative element, the -I inductive effect causes the hydrogen to become less tightly held by the C-Cl bond.

As the number of electron withdrawing substituents increaseses, the acid ionizes much more and becomes stronger.CH3OH < FCH2OH < F2CHOH < F3COH

Hence, the order of decreasing acid strength is;

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4. What is the molar mass of CuSO4?
IceJOKER [234]

Answer: 159,609 g/mol

Explanation:

8 0
3 years ago
Read 2 more answers
Ion Concentration 1. Which solution has the greatest [SO42-]: a) 0.075 M H2SO4 b) 0.15 M Na2SO4 c) 0.080 M Al2(SO4)3?
soldi70 [24.7K]

Answer:

c) 0.080 M Al₂(SO₄)₃

Explanation:

Ion [SO₄²⁻] concentration of each solution is:

a) 0.075 M H₂SO₄: <em>[SO₄²⁻] = 0.075M</em>. Because 1 mole of H₂SO₄ contains 1 mole of SO₄²⁻

b) 0.15 M Na₂SO₄: <em>[SO₄²⁻] = 0.15M</em>. Also, 1 mole of Na₂SO₄ contains 1 mole of SO₄²⁻

c) 0.080 M Al₂(SO₄)₃ [SO₄²⁻] = 0.080Mₓ3 =<em> 0.240M</em>. Because 1 mole of Al₂(SO₄)₃ contains 3 moles of SO₄²⁻.

<h3>Thus, the soluion that has the greatest [SO₄²⁻] is 0.080 M Al₂(SO₄)₃</h3>
8 0
3 years ago
Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en
Bezzdna [24]

Answer:

  • 10.19 g CO₂
  • 4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

  • C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:

  • 1 gallon = 3.785 L
  • 1 L = 1000 mL

Now we<u> convert 1.24 gallons to mL</u>:

  • 1.24 gallon * \frac{3.785L}{1gallon} *\frac{1000mL}{1L} = 4693.4 mL

We <u>calculate the mass of Octane</u>:

  • 4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:

  • CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * \frac{8molCO_{2}}{1molOctane} * 44 g/mol =  10.19 g CO₂
  • H₂O ⇒ 3.30 g Octane ÷ 114g/mol * \frac{9molH_{2}O}{1molOctane} * 18 g/mol = 4.69 g H₂O

7 0
4 years ago
A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what
Charra [1.4K]
1) mass composition

N: 30.45%
O: 69.55%
   -----------
   100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

=> N2O4

Answer: N2O4
3 0
3 years ago
H3PO4 + Ca(OH)2 → Ca(H2PO4)2 + H2O
aniked [119]

Given question is incomplete. The complete question is as follows.

Balance the following equation:

H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + H_2O

Answer: The balanced chemical equation is as follows.

2H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + 2H_2O

Explanation:

When a chemical equation contains same number of atoms on both reactant and product side then this equation is known as balanced equation.

For example, H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + H_2O

Number of atoms on reactant side:

H = 5

P = 1

O = 6

Ca = 1

Number of atoms on product side:

H = 6

P = 2

O = 9

Ca = 1

In order to balance this equation, we will multiply H_3PO_4 by 2 on reactant side and we will multiply H_2O by 2 on product side. Hence, the balanced chemical equation is as follows.

2H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + 2H_2O

8 0
3 years ago
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