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user100 [1]
2 years ago
12

Potential and kinetic energy are similar in that -COME ON BESTIES I NEED HELP​

Chemistry
2 answers:
Stels [109]2 years ago
7 0

Answer:

I'm not 100% but a possible answer could be: No energy could either be created or destroyed

LMK if not but good luck!

Masja [62]2 years ago
3 0

Answer:

They refer to energy that moves. Kinetic energy is energy that is currently moving. Potential refers to energy that has yet to move, or simply energy in wait

Explanation:

You might be interested in
Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

3 0
3 years ago
Why does it take the truck longer to stop than the car?
Ne4ueva [31]

Answer:

Newton's 1st Law.

4 0
3 years ago
If an ideal gas at a constant temperature is initially at a pressure of 3.8 atm and is then allowed to expand to a volume of 5.6
BabaBlast [244]

Answer:

The most common example is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.If an ideal gas at a constant temperature is initially at a pressure of 3.8 atm and is then allowed to expand to a volume of 5.6 L and a pressure of 2.1 - 18914… ... of 5.6 L and a pressure of 2.1 atm, what is the initial volume of the gas? ... An ideal gas is at a pressure of 1.4 atm and has a volume of 3 L.

Explanation:

I hope I help :)

7 0
2 years ago
Drawing Conclusions
Dmitrij [34]

<u>Answer:</u>

<em>The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock. </em>

<u>Explanation:</u>

Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.

The time elapsed will be log (L) atoms X half-life.

4 0
3 years ago
Determine the type of alcohol corresponding to each given description or name. 1-pentanol 3-ethyl-3-pentanol 2-hexanol An alcoho
pochemuha

Answer:

1). 1-pentanol - <u>Primary</u>

2). 3-ethyl-3-pentanol - <u>Tertiary</u>

3). 2-hexanol - <u>Secondary</u>

4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - <u>Secondary</u>

5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - <u>Primary</u>

6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - <u>Tertiary</u>

Explanation:

The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.

In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.

In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.

In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example:  3-ethyl-3-pentanol, -tert -butyl alcohol, etc.

7 0
2 years ago
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