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4 years ago

Jane is sliding down a slide. What kind of motion is she demonstrating?

Chemistry

1 answer:

Alex Ar [27]4 years ago

8 0

Im pretty sure it would be kinetic

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What is the molarity of a solution that contains 0.400 mol HCI in 9.79 L solutions

USPshnik [31]

Molarity's formula is known as: Molarity(M)=moles of solute/liters solution.

In this case we are already given moles and liters so you just have to plug the numbers into the equation.

0.400 mol HCL/9.79L solution=0.040858M

If you were to use scientific notation, the answer will be: 4.1*10^-2, but otherwise, you can just use the decimals above and round appropriately as you see fit.

6 0

3 years ago

How do the particles in each state of matter move

hodyreva [135]

the particles of solids move but very slowly.

the particles of liquids move moderately fast

the particles of gas move very fast.

The reasons for this movement is the space the particles are together. Since there is not move space between solids, the particles move slowly, water there is moderate space, and air there is a lot of space.

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3 years ago

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igor_vitrenko [27]

Answer:

I don’t know

Explanation:

5 0

3 years ago

Read 2 more answers

A 42.0 mL sample of an NaOH solution is placed in a flask with a few drops of phenolphthalein (indicator). If 11.7 mL of a 0.325

olganol [36]

Please like if you get the answer :)

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2 years ago

A 29.7 g sample of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to

Softa [21]

Answer: The mass of iron in the ore is 10.9 g

Explanation:

We are given:

Mass of iron (III) oxide = 15.6 g

We know that:

Molar mass of Iron (III) oxide = 159.69 g/mol

Molar mass of iron atom = 55.85 g/mol

As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.

To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:

In 159.69 g of iron (III) oxide, mass of iron present is $(2\times 55.85)=111.7g$

So, in 15.6 g of iron (III) oxide, mass of iron present will be = $\frac{111.7g}{159.69g}\times 15.6g=10.9g$

Hence, the mass of iron in the ore is 10.9 g

7 0

3 years ago