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2 years ago

How many moles of MgCl2 are there in 302 g of the compound?

Physics

2 answers:

VashaNatasha [74]2 years ago

8 0

MgCl2 molar mass is about 94g/mol so 302/94 is about 3.2. So the answer is 3.2 moles in 302 g.

Hope this helps:)

lisov135 [29]2 years ago

3 0

Answer:

‍♀️‍♀️‍♀️‍♀️sorrryyyyy

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What is the displacement of the student from the bus stop to the mailbox?

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Answer:C

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2 years ago

A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki

Mars2501 [29]

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

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$\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s$

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2 years ago

Arunning back practices for his upcoming football game by running drills. He runs forward 62.4 yards, and runs backward for

Mama L [17]

Answer: He has only move 0.2 yards

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3 years ago

a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction 30° east of north

liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement $D_{1}$ = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey $D_{2}$ = 20 km (red arrow) is at an angle with reference to y-axis.

So, we first find the components of the red arrow along X and Y.

Component of $D_{2}$ along X-axis is given by $D_{2x} = D_{2} Sin 30$ = 10 km

Component of $D_{2}$ along Y-axis is given by $D_{2y} = D_{2} Cos 30$ = 17.32 km

We now add all the vectors along X and along Y separately.

Net Displacement along X $D_{netX} = 10 km$

Net Displacement along Y $D_{netY} = 30 + 17.32$ = 47.32 km

Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the $D_{net}$

$D_{net} = \sqrt{D_{netX} ^{2} + D_{netY} ^{2}}$

Therefore, [tex]D_{net} = 48.37 km.

Hence, the distance between the ports P and R is 48 km.

6 0

3 years ago