Which is an example of kinetic energy?

Many chemistry problems result in equations of the form

Gnom [1K]

Answer:

When this equation is solved, the two values of the unknown are 0.0643 and -0.082

Explanation:

Given

$1.77 * 10^{-2} = \frac{x^2}{0.298 - x}$ --- the actual equation

Required

The values of x

We have:

$1.77 * 10^{-2} = \frac{x^2}{0.298 - x}$

Cross Multiply

$1.77 * 10^{-2} * (0.298 - x)= x^2$

Multiply both sides by 100

$1.77 * (0.298 - x)= 100x^2$

Open bracket

$0.52746 - 1.77x= 100x^2$

Rewrite as:

$100x^2 + 1.77x - 0.52746 =0$

Using quadratic formula:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where:

$a = 100; b = 1.77; c = -0.52746$

So, we have:

$x = \frac{-1.77 \± \sqrt{1.77^2 - 4*100*- 0.52746 }}{2*100}$

$x = \frac{-1.77 \± \sqrt{214.1169}}{2*100}$

$x = \frac{-1.77 \± 14.63}{200}$

Split

$x = \frac{-1.77 + 14.63}{200}\ or\ x = \frac{-1.77 - 14.63}{200}$

$x = \frac{12.86}{200}\ or\ x = \frac{-16.40}{200}$

$x = 0.0643\ or\ x = -0.082$

3 0

3 years ago

A sample of gas occupies 3.00 L with 5.00 moles present. What would

pantera1 [17]

3/5 times 5/3x = 8*3/5. X=24/5 simplified would be x= 4.8 L.

5 0

3 years ago

A motorist discovers that her auto tire measures 29.2 psi rather than the recommended 35 psi. If the tire’s volume is assumed co

I am Lyosha [343]

Answer:

0.20 moles

Explanation:

The pressure is proportional to the quantity of gas at a given temperature and volume. So, the quantity needs to be increased by a factor of ...

(35 psi)/(29.2 psi) = 175/146 ≈ 1.19863

The fractional increase required is ...

1.19863 -1 = 0.19863

__

The quantity of air currently in the tire is ...

1 mol·519.67°R/(atm·23.6442 L) × (29.2/14.7 atm) × (11.6 L) / (45+459.67)°R

= 1.0035 mol

so we need to add ...

(fraction to add) × (current quantity) = amount to add

0.19863 × 1.0035 mol = 0.1993 mol = amount to add

About 0.20 moles of air must be added to the tire to bring the pressure up.

5 0

3 years ago

When 25.0 ml of 0.500 m h2so4 is added to 25.0 ml of 1.00 m koh in a coffee-cup calorimeter at 23.50°c, the temperature rises to

AlladinOne [14]

The chemical reaction that occurs between the given substances is a neutralization reaction as shown below :

H_2SO_4 + 2KOH -> K_2SO_4 + 2H_2O

1 mol 2 mol 1 mol 2 mol

The number of moles of the given substances is calculated as shown :

Number of moles of H_2SO_4 = 25.0 mL x 0.50 M = 12.5 millimoles

Number of moles of KOH = 25.0 mL x 1.00 M = 25.0 millimoles

As 1 mol of sulfuric acid reacts with 2 mol of KOH to give 2 mol of water, 12.5 millimoles of sulfuric acid completely reacts with 25.0 millimoles of KOH to give 25.0 millimoles of water.

Total volume of the solution = 25.0 mL + 25.0 mL = 50.0 mL.

Density of water is 1 g/mL. Use this to calculate the mass of the solution.

Mass of the solution – 50.0 mL x 1 g/mL = 50.0

The specific heat of water is 4.184 J/gK. The temperature of the solution is increased from 23.5 degrees Celsius to 30.17 degrees Celsius.

The amount of heat released = 4.184 J/gK x 50.0 g x (30.17C – 23.50C) 1395 J

The amount of heat released per one mole of water formed can be calculated as shown :

The amount of heat released for formation of mole water = 1395 J / (25.0 m mol x 1mol/1000 m mol)

= 55,800 J

8 0

3 years ago

What do these two changes have in common?

Vladimir79 [104]

They are both physical changes (change in physical structure/appearance)

hope this helps :)

6 0

3 years ago