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Umnica [9.8K]
3 years ago
10

A ball is sitting at the top of a ramp. As the ball rolls down the ramp, the potential energy of the ball decreases. What

Chemistry
1 answer:
Ivenika [448]3 years ago
5 0

It is converted to kinetic energy since it is now moving.

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Select all of the statements that are true about a buffer solution. Select one or more: a. A buffer solution reacts with basic s
zhuklara [117]

Answer:

a. A buffer solution reacts with basic solutions.

c. A buffer solution reacts with acidic solutions.

e. A buffer solution resists small changes in pH

Explanation:

1. Buffer questions

a, c, and e are TRUE. A buffer resists a change in the pH when small amounts of a strong acid or base are added to it.

b is wrong. A buffer can have a pH of 7, but it can also have many other pH values.

d is wrong. Most buffers are colourless, and they resist a change in pH.

2. Titration curves

The solution is the best buffer at the mid-point of the titration curve.

In the figure below, the equivalence point is at 13 mL, so the mid-point is at 6.5 mL.  

The solution is buffered at pH 3.2.

However, the solution is a buffer at any point in the range pH = 3.2 ± 1.

That would be in the range of 1 mL to 12 mL.

The buffering ability becomes worse the further you are from the mid-point of the titration.

5 0
3 years ago
7. A figure skater skating across ice, who grabs another skater and brings him along the ice with her, is an example of what typ
xenn [34]

Answer:

The type of collision is A. Inelastic collision.

Explanation:

The macroscopic collisions are generally inelastic and do not conserve the kinetic energy, although of course the total energy is conserved. The inelastic collision is one in which the objects that collide remain together after the collision.

So, a figure skater skating across ice, who grabs another skater and brings him along the ice with her is a clear example of inelastic collision.

6 0
3 years ago
Read 2 more answers
Why all samples of a given substance have the same intensive properties?
Karolina [17]

Answer:

Every sample of a given substance has identical intensive properties because every sample has the same composition

4 0
3 years ago
YOU DO:
mihalych1998 [28]

Answer:

67.1%

Explanation:

Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:

<em>Moles Na₂CO₃ - 105.99g/mol-:</em>

6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.

As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:

0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃

And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):

0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.

And percent of NaHCO₃ in the sample is:

10.06g NaHCO₃ / 15g Sample * 100 =

<h3>67.1%</h3>
7 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
2 years ago
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