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3 years ago

7. A cyclist starting from rest accelerates at a rate of 6 m/s until she reaches 30 m/s. How long

Chemistry

1 answer:

Hunter-Best [27]3 years ago

4 0

Answer:

Explanation:

The given values are:

a= 6m/s²

u= 0 m/s

v = 30 m/s

t= ?

The formula is :

$a=\frac{v-u}{t}$

thus,

$t=\frac{v-u}{a}\\t=\frac{30-0}{6}\\t= 5 seconds$

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While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e

sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Explanation:

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

Initial: 1.8 4.7

At eqllm: 1.8-x 4.7-x

Evaluating the value of 'x'

$\Rightarrow (1.8-x)=1.16\\\\x=0.64$

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}$

$p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm$

Putting values in above expression, we get:

$K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136$

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

Initial: 2.36 4.06 0.64

At eqllm: 2.36-x 4.06-x 0.64+x

Putting value in the equilibrium constant expression, we get:

$0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41$

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0

3 years ago

A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t

Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 2 atm
T1= 50 C= 323 K (being 0 C= 273 K)
P2= 3.2 atm
T2= ?

Replacing:

$\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}$

Solving:

$T2*\frac{2 atm}{323 K} =3.2 atm$

$T2=3.2 atm*\frac{323 K}{2 atm}$

T2= 516.8 K= 243.8 C

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

5 0

3 years ago

In a chemical reaction between copper metal and silver nitrate, 12.7 g of copper(II) and 64.5 g of silver nitrate react, and 38.

hodyreva [135]

Answer:What should you do if you realize during research that your original theory is wrong

Explanation:

8 0

3 years ago

PLEASE HELP! See picture

Darya [45]

Answer:

See Explanation

Explanation:

The equation of the reaction;

KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)

Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles

Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.

Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles

Since the reaction is 1:1, 0.45 moles of K2SO4 is produced

Hence K2SO4 is the limiting reactant.

Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g

So;

1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles

Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH

5 0

3 years ago

Justin mixed two clear liquids together during an experiment and made the following observations:

svp [43]

Answer:

B. A chemical change occurred which caused the liquid's physical properties to change.

Explanation:

The reduction of the temperature of the system meant that the reaction absorbed heat energy from it. This shows that a chemical reaction was in progress. New products were formed, and this is proved by the change in the color to blue.

6 0

3 years ago

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