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Masteriza [31]
3 years ago
11

Which of these conjugate acid-base pairs will not function as a buffer?A) HNO3 and NO3^-B) HCO3^- and CO3^2-C) C2H5COOH and C2H5

COO^-
Chemistry
1 answer:
marin [14]3 years ago
5 0
<h3><u>Answer;</u></h3>

A) HNO3 and NO3^-

<h3><u>Explanation;</u></h3>
  • <em><u>HNO3 is a strong  acid and NO3 is its conjugate base, meaning it will not have any tendency to withdraw H+ from solution.</u></em>
  • Buffers are often prepared by mixing a weak acid or base with a salt of that weak acid or base.
  • The buffers resist changes in pH since they contain acids to neutralize OH- and a base to neutralize H+. Acid and base can not consume each other in neutralization reaction.
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What would happen if species I'm your area disappeared? What might happen?
zhannawk [14.2K]
The population will get low because the food we eat is animals which is species then the animals that eat that prey will die off because that might be there only prey....then if more species is gone there will be nothing left

I Hope this helped
7 0
4 years ago
Which of the following is NOT the property of a non-metal? A. Non-rigid structure B. Not ductile C. Low density D. Malleable
Vedmedyk [2.9K]

Answer:

The answer is D.

Explanation:

Malleable is the property of <u>m</u><u>e</u><u>t</u><u>a</u><u>l</u> not non-metal.

4 0
4 years ago
Read 2 more answers
3. In the apparatus below, the seawater is an example of a(n) ——
san4es73 [151]

In the apparatus of the electrochemical cell, the seawater is an example of an electrolyte.

<h3>What is an electrolyte?</h3>

An electrolyte is a component of the electrochemical cell that is a solution of solvent or water that contains dissolved ions.

The salts of sodium, chloride, and potassium are the main components of seawater and a great electrolyte.

Therefore seawater is an example of an electrolyte.

Learn more about electrolytes here:

brainly.com/question/2285692

#SPJ1

3 0
2 years ago
In a study designed to prepare new gasoline-resistant coatings, a polymer chemist dissolves 6.143 g poly(vinyl alcohol) in enoug
Ghella [55]

Answer:

MM = 5,521.54 g/mol

Explanation:

To solve this, we need to use the expression for osmotic pressure which is the following:

π = MRT (1)

Where:

M: Concentration of the solution

R: gas constant (0.082 L atm/ mol K

T: temperature in K

25 °C in Kelvin is: 25 + 273.15 = 298.15 K

Now, we do not have the concentration of the solution, but we do have the mass. and the concentration can be expressed in terms of mass, molar mass and volume:

Concentration (M) is:

M = n/V (2)

and n (moles) is:

n = m/MM (3)

Therefore, if we replace (2) and (3) in (1) we have:

π = mRT/V*MM

Solving for MM we have:

MM = mRT/πV (4)

All we have to do now, is replace the given data and we should get the value of the molar mass:

MM = 6.143 * 0.082 * 298.15 / 0.1 * 0.272

MM = 150.1859 / 0.0272

<em>MM = 5,521.54 g/mol</em>

<em>This is the molar mass.</em>

7 0
3 years ago
What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressure depression of 0.734 mmHg a
yaroslaw [1]

<u>Answer:</u> The Van't Hoff factor for KCl is 1.74

<u>Explanation:</u>

We are given:

Molality of solution = 1 m

This means that 1 mole of a solute is present in 1 kg of solvent (water) or 1000 grams of water

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of solute = 1 moles

Total moles = [1 + 55.56] = 56.56 moles

Putting values in above equation, we get:

\chi_{(solute)}=\frac{1}{56.56}=0.0177

The equation used to calculate relative lowering of vapor pressure follows:

\frac{p^o-p_s}{p^o}=i\times \chi_{solute}

where,

\frac{p^o-p_s}{p^o} = relative lowering in vapor pressure = 0.734 mmHg

i = Van't Hoff factor = ?

\chi_{solute} = mole fraction of solute = 0.0177

p^o = vapor pressure of pure water = 23.76 torr

Putting values in above equation, we get:

\frac{0.734}{23.76}=i\times 0.0177\\\\i=1.74

Hence, the Van't Hoff factor for KCl is 1.74

7 0
3 years ago
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