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1 year ago

1. A crane lifts a 75kg mass a height of 8 m. Calculate the gravitational potential energy

Chemistry

1 answer:

koban [17]1 year ago

6 0

Answer:

GP.E = 5880 j

Explanation:

Given data:

Mass = 75 kg

height = 8 m

Potential energy = ?

Solution:

The formula for gravitational potential energy is

GPE = mgh

m = mass in kilogram

g = acceleration due to gravity

h = height in meter above the ground

Formula:

GP.E = mgh

Now we will put the values in formula.

g = 9.8 m/s²

GP.E = 75 Kg × 9.8 m/s²× 8 m

GP.E = 5880 Kg.m²/s²

Kg.m²/s² = j

GP.E = 5880 j

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Consider the following hypothetical reaction: 2 P + Q â†’ 2 R + S The following mechanism is proposed for this reaction: P + P Q

aleksandr82 [10.1K]

Answer: the answer is option (D). k[P]²[Q]

Explanation:

first of all, let us consider the reaction from the question;

2P + Q → 2R + S

and the reaction mechanism for the above reaction given thus,

P + P ⇄ T (fast)

Q + T → R + U (slow)

U → R + S (fast)

we would be applying the Rate law to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

It is important to understand that laws based on experiment do not allow for intermediate concentration.

The mechanism steps for the reactions in the question are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.

adding this steps gives a final overall reaction reaction.

2P + Q ------------˃ 2R + S

Thus the rate equation is given as

Rate (R) = K[P]²[Q]

cheers, i hope this helps

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2 years ago

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:- The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$

= 30.0 L $C_2H_6$

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

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Answer:

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