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3 years ago

How many moles of MgCl2 are there in 315 g of the compound? mol

Physics

2 answers:

Nikitich [7]3 years ago

8 0

6.022m molecules in one gram

jeka943 years ago

4 0

315g/95gmol-1
3.315 moles of MgCl2

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A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0

oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box $m=2\ kg$

speed of box $v=1.9\ m/s$

distance moved by the box $x=10\ cm$

coefficient of kinetic friction $\mu _k=0.66$

Friction force $f_r=\mu_kN$

$f_r=0.66\times mg$

$f_r=0.66\times 2\times 9.8=12.936 \N$

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

$\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2$

$\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2$

$3.61-1.2936=0.005\times k$

$k=463.28\ N/m$

3 0

3 years ago

Read 2 more answers

The average diameter of one tennis ball in a package of three is 6.8 cm. Which of the following is the combined volume of all th

Crank

We want to find the combined volume of 3 tennis balls. We will get that the combined volume is 493.7 cm^3

First, remember that for a sphere of diameter D, the volume is:

$V = \frac{4}{3}*3.14*(\frac{D}{2})^3$

Where 3.14 is pi.

Here we know that the average diameter of a tennis ball is 6.8cm, then we can replace that in the above equation to find the volume (in average) of a single tennis ball:

$V = \frac{4}{3}*3.14*(\frac{6.8cm}{2})^3 = 164.5 cm^3$

Now, in 3 balls of tennis, the combined volume will be 3 times the above one, this is:

$3*V = 3*164.5cm^3 = 493.7 cm^3$

If you want to learn more about volumes, you can read:

brainly.com/question/10171109

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2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

4 years ago

A rocket will move upward as long as which condition applies?

Dennis_Churaev [7]

The force of thrust is greater than the force if gravity !
Answer found on quizlet !

8 0

3 years ago

If you exert a force of 100.0 N to lift a box a distance of 0.5 m, how much work do you do? 200 J 400 J 50 J 26 J

jeyben [28]

Work = force x distance
= 100N (force) x 0.5m (distance)
= 50J

4 0

3 years ago