In an experiment, 1 mol of propane is burned to form carbon dioxide and water.

Which describes newton’s law of universal gravitation?

Zielflug [23.3K]

(B) All objects attract other objects

To be specific, the following formula defines this theory very clearly:

F = G * (m1 * m2) / r^2

The Force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

8 0

3 years ago

What is the density of 53.4 wt queous naoh solution if 16.7 ml of the solution diluted to 2.00 l gives 0.169 m naoh?

vampirchik [111]

The density of 53.4 wt aqueous NaOH solution is 0.809 g/ml

Given data:

The mass percent of NaOH is 53.4.
Volume of NaOH diluted is 16.7 ml.
The volume of diluted solution is 2.00 L =2000 ml.
Concentration of diluted solution is 0.169 M.

First, we find the initial concentration of NaOH by using the following formulae,

M₁V₁ = M₂V₂

Where,

M₁ is the initial molarity of NaOH

M₂ is the molarity after dilution

V₁ is the initial volume of NaOH

V₂ is the final volume after dilution.

Substituting the values,

M₁ × 16.7 ml = 0.169 M × 2000 ml,

M₁ = $\frac{0.169 M *2000 ml}{16.7 ml}$

M₁ = 20.2 M.

Thus, the initial concentration of NaOH is 20.2 M.

we know, 1 M solution contains 1 mol of substance present in 1 L solution,

Thus, 20.2 M solution will have 20.2 mols of NaOH.

Now, we can find the mass of NaOH by using the number of moles and molar mass.

molar mass of NaOH is 40 g/mol.

Mass = no. of moles × molar mass

= 20.2 mol × 40 g/mol

= 808 g.

Thus, the mass of NaOH is 808g.

53.4 wt of NaOH means 53.4 g of NaOH in a 100 g solution,

Thus, 808 g of NaOH will be present in ,

⇒ $\frac{53.4 g NaOH}{100 g solution} = \frac{808 g NaOH}{x g solution}$

⇒ 1513.1 g

Now, Convert the grams of NaOH to milliliters, using the density of NaOH at room temperature.

The density of NaOH at room temperature is 1.515 g/ml,

Density = $\frac{mass}{volume}$

⇒ 1.515 g/mol = $\frac{1513.1 g}{volume}$

⇒ volume = $\frac{1513.1 g}{1.515 g/mol}$

⇒ volume = 998.7 ml.

Thus, the volume of NaOH is 998.7 ml.

Hence, we know,

the mass of NaOH is 808 g
the volume of NaOH is 998.7 ml

Substituting the values,

Density = 808 g / 998.7 g/ml

⇒ Density = 0.809 g/ml

Thus, the density of 53.4 wt aqueous NaOH is 0.809 g/ml.

To learn more about Density here

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3 0

2 years ago

Determine the centripetal force on a vehicle rounding a circular curve with a radius of 80 m at a constant speed of 90 km/h if t

quester [9]

Mass of the vehicle = 2000 kg
Velocity of the vehicle = 90 km/hr
= 25 m/s
Radius of the curve = 80 m
Then
Centripetal force = [2000 * (25)^2]/80
= (2000 * 625)/80
= 15625 kg m/s
= 15625 Newton second
I hope that this is the answer that you were looking for and the answer has come to your desired help.

3 0

3 years ago

Which of the following would increase the output force of a lever?

Vedmedyk [2.9K]

A. Increase the distance between the effort and the fulcrum

5 0

3 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

1. Part A: 648N
2. Part B: 324J
3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

3. Part C

What maximum force must you apply to move the platform to the position in Part B?

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

8 0

3 years ago