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Ivan
3 years ago
9

Consider the car driving around a circular track .Once stated,the driver maintains a speed 0f 45m/h.Compare the driver's velocit

y with his acceleration in this case.
A.)The car is accelerating because it is changing direction but the velocity is constant.
B.)The car is traveling at a constant rate of speed so it is not accelerating and the velocity does not change
C.)Acceleration is the change in velocity so the car is experiencing a change in velocity and it is accelerating
D.)The car is traveling at a constant rate of speed so it is not accelerating but the velocity changes because the direction the car travels is constantly changing.
Chemistry
1 answer:
Lyrx [107]3 years ago
6 0
A i hope that helped i love you guys have a great day
And merry chrismas

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Part A. Two containers, one at 305 K and the other at 295 K, are placed in contact with each other. 1. 1 J of heat flows from th
posledela

Answer:

0.00011 JK.

The process does NOT violate the second law of thermodynamics

Explanation:

The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.

=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.

The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.

The change in entropy of the cold container = 1/295 = 0.00339 J/K.

Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.

Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.

The process does NOT violate the second law of thermodynamics.

7 0
3 years ago
9A. A sample of hydrogen at 1.56 atm had it's pressure decreased to 0.73
densk [106]

Answer:

351.43mL

Explanation:

To calculate the original volume of hydrogen gas in this question, the Boyle's law equation will be used. Boyle's law equation is:

P1V1 = P2V2

Where; P1 = initial pressure

V1 = initial volume

P2 = final pressure

V2 = final volume

According to this question, the P1= 1.56atm, V1 = ?, P2 = 0.73atm, V2 = 751mL

Hence;

P1V1 = P2V2

1.56 × V1 = 0.73 × 751

1.56 V1 = 548.23

V1 = 548.23/1.56

V1 = 351.43mL

Therefore, the original volume of hydrogen gas is 351.43 mL.

4 0
3 years ago
What is the mass in grams of 0.687 moles of NH three
stepan [7]
NH3 = 14 +3*1=17 g/mol
Mass = mol * molar mass = 0.687 * 17= 11.679 g
Can u mark it brainliest?
3 0
3 years ago
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug
OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 =  log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] <em>(1)</em>

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = <em>0.035M</em>

Thus: [CH₃COO⁻] = 0.100M - 0.035M = <em>0.065M</em>

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles +  2.646x10⁻³ moles = <em>8.071x10⁻³ moles of CH₃COOH</em>. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = <em>0.296</em>

4 0
3 years ago
How many grams of sodium acetate are in solution in the third beaker?
Kipish [7]

Answer:

46g of sodium acetate.

Explanation:

The data is: <em>Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156g of the salt in 100 mL of water at 100°C and then slowly cooling it to 20°C. Because the solubility of sodium acetate in water at 20°C is 46g per 100mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.</em>

The third solution is the result of the equilibrium in the solution at 20°C. As the maximum quantity that water can dissolve of sodium acetate at this temperature is 46g per 100mL and the solution has 100mL <em>there are 46g of sodium acetate in solution. </em>The other sodium acetate precipitate because of decreasing of temperature.

I hope it helps!

6 0
3 years ago
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