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Illusion [34]
3 years ago
5

Which of the following claims about a binary compound composed of elements with the same electronegativity is most likely to be

true?
A
Tully compound has properties similar to those of both elements.
The bonding in the compound is nonpolar covalent.
с
The boiling point of the compound is above 1000°C.
D
The compound contains strong ionic bonds.
Chemistry
1 answer:
Mazyrski [523]3 years ago
7 0

I think it's D

Hope that helps

You might be interested in
The volume of a gas is 550 mL at 960 mm Hg and 200.0 C. What volume
Masja [62]

Answer:

The volume will be 568.89 mL.

Explanation:

Boyle's law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, the pressure of the gas is directly proportional to its temperature. Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Where P = pressure, T = temperature, K = Constant

Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

\frac{V}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Studying an initial state 1 and a final state 2, it is fulfilled:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

In this case:

  • P1= 960 mmHg
  • V1= 550 mL
  • T1= 200 C= 473 K (being 0 C=273 K)
  • P2= 830 mmHg
  • V2= ?
  • T2= 150 C= 423 K

Replacing:

\frac{960 mmHg*550 mL}{473K} =\frac{830 mmHg*V2}{423 K}

Solving:

V2=\frac{423 K}{830 mmHg} *\frac{960 mmHg*550 mL}{473K}

V2= 568.9 mL

<u><em>The volume will be 568.89 mL.</em></u>

4 0
3 years ago
Complete and balance each equation. If no reaction occurs write N/R.<br> NaOH(aq) + FeBr3(aq)
Shtirlitz [24]
FeBr3(aq) + NaOH(aq) = FeOH(s) + NaBr3(aq) 
8 0
3 years ago
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic
Mademuasel [1]

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>

<em />

When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

<em>Where X is reaction coordinate.</em>

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

<h3>1.33%</h3>

<em />

<em />

<em />

4 0
3 years ago
A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region o
Llana [10]

Answer:

5.75

Explanation:

Weak acid and strong base,

Make it a two part problem

The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH

1.Parameters

HC2H3O2 =10mL x 0.75M = 7.5 mmol

NaOH =22mL x 0.30M = 6.6 mmol

R HC2H3O2 + OH- --> C2H3O2- + H2O

I 7.5 mmol 6.6 mmol 0 mmol ignore

C -6.6 mmol -6.6 mmol +6.6 mmol

E 0.9 mmol 0.0 mmol 6.6 mmol

2.) HC2H3O2 --> H+ + C2H3O2-

Initial concentrations

[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M

[H+]:

[C2H3O2-]: 6.6 mmol x 32mL = 0.206M

Concentrations at equilibrium

[HC2H3O2]: 0.0281M - x

[H+]: [C2H3O2-]: 0.206M + x

If x is small and can be ignored except [H+]

Substitute and solve

1.3 x 10-5 = (x)(0.206)

(0.0281)

X= 1.77 x 10-6M => pH = 5.75

3 0
3 years ago
1. A 5.455-g sample of impure CaCl2 is dissolved and treated with excess potassium carbonate solution. The dried CaCO3 (calcium
Helen [10]

Answer:81.6%

Explanation:

Mass of CaCO3=4.010 g

Molar mass of CaCO3= 40+12+(16×3) = 100 g/mol.

Recall: number of moles(n)= mass÷ molar mass.

n=4.010÷100 = 0.0401 mol.

Molar mass of CaCl2 = 40+71= 111 g/mol.

Number of mol of CaCl2 = 5.455÷111= 0.04914 g/mol.

Mass of CaCl2 = 0.0401 × 111 = 4.4511 g of CaCl2.

Percent by mass of CaCl2 = (4.4511÷5.455) × 100

= 0.815967 ×100 = 81.5967%

Approximately; 81.6%.

8 0
4 years ago
Read 2 more answers
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