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Anettt [7]
2 years ago
11

If a small amount of the unknown solute fails to dissolve in the lauric acid, will the molar mass that you calculate for unknown

solute be too high or too low?
Chemistry
1 answer:
dimaraw [331]2 years ago
7 0
If the solute fails to dissolve in the lauric acid, the molar mass that can be calculated based on the concentration of the solute dissolved in the lauric acid would be too low. This is because there are unaccounted masses that cannot be included in the calculation. This is a case of underestimation of the molar mass.

To avoid that this happens, ensure that all of the solute dissolves well in solvent by mechanical means (stirring, shaking, etc) or introducing heat. 
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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
marin [14]
<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.
5 0
3 years ago
Hess’s law
Delvig [45]

From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ

<h3>What is Hess' law?</h3>

Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.

From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:

A---> C = A ---> B + B ---> C

ΔH of A---> C = 30 kJ + 60 kJ

ΔH = 90 kJ

Therefore, the enthalpy of the reaction A---> C is +90 kJ

The above reaction A---> C can be shown in the enthalpy diagram below:

A -------------------> C (ΔH = +90 kJ)

\ /

\ / (ΔH = +60 kJ)

(ΔH = +30 J) \ /

> B

Learn more about enthalpy and Hess law at: brainly.com/question/9328637

8 0
2 years ago
How many moles of helium are 8.84×1024 atoms of He?
insens350 [35]

Answer:

14.68 moles of He

Explanation:

To do this, just remember Avogadro's Constant or Avogadro's number. This constant tells us how many units ( in this case atoms) there are in a mole of ANY type of substance.

Avogadro's constant is 6.022140857 × 10²³ units per mole.

Now that we know how many atoms there are in 1 mole, we can use this as our conversion factor.

8.84 x 10²⁴ atoms of He →  moles of He

8.84\times10^{24} atoms of He\times\dfrac{1moleofHe}{6.022140857\times10^{23}atomsofHe}=14.68molesofHe

So the answer would be:

14.68 moles of He

4 0
2 years ago
When multiplying numbers in scientific notation what do you do with exponents?
Basile [38]

Answer:

Subtract them.

Explanation:

''''"Since all number in scientific notation have base 10, we can always multiply them and divide them. To multiply two numbers in scientific notation, multiply their coefficients and add their exponents. To divide two numbers in scientific notation, divide their coefficients and subtract their exponents."""""

I was actually learned about this in school just found an source.

6 0
3 years ago
Which compound is an Arrhenius base?
Dmitry_Shevchenko [17]
I was hoping that some choices would be given to choose from. As there are no choices given, so i am answering the question based on my knowledge and hope that it comes to your help. Calcium hydroxide is a good example of Arrhenius base. An Arrhenius base is actually a substance that releases a hydroxyl ion in water.
3 0
3 years ago
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