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kifflom [539]
3 years ago
10

What is known about a reaction with a negative enthalpy?

Chemistry
1 answer:
Debora [2.8K]3 years ago
3 0

Answer:

B.the reaction is spontaneous

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You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration
vovangra [49]

<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM



7 0
3 years ago
Which of these elements is unlikely to have a reaction with any element or compound?
Kay [80]
Argon is a noble gas. Argon has a full outer shell. This makes it so that it does not need to react with any of the other elements to be stable.


With Rubidium and Cobalt its a whole different story.

I hope that helps!
8 0
3 years ago
What was the original element formed moments after the Big Bang? What then created higher order elements?
Semmy [17]
Hydrogen (H) was first, followed by helium (He).
6 0
3 years ago
Use Boyle's Law to explain what would happen to the volume of a helium- filled balloon if it was carried underwater by a diver t
oksano4ka [1.4K]

The deeper the diver takes the helium balloon, the more it reduces in size. This is due to the pressure of the water column above pressing on the balloon. According to Boyle’s law (P= k*1/V.), as the volume of the balloon decreases, the pressure of the helium inside increases.

4 0
3 years ago
Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s)
klemol [59]

Answer:

a) 2.01 g

Explanation:

  • Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:

  • 0.0302 mol AgNO₃ * \frac{1molNa_2CO_3}{2molAgNO_3}  = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

  • 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:

  • 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).

8 0
3 years ago
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