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kirill [66]
3 years ago
9

The different phases of the Moon refer to A. Changes in the Moon's shape. B. Changes in the Moon's visibility. C. Changes in sea

sons on the Moon. D. Changes in the Moon's rotation.
Chemistry
1 answer:
Sav [38]3 years ago
5 0

Answer:

B. Changes in the Moon's visibility.

Explanation:

The moon's shifting of position makes it "appear" as being smaller or bigger. This is simply because the moon is rotating and the sun's light shines on a certain area of the moon, and depending on where we are on the Earth is how the moon looks or appears to our perspective.

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What is the difference between applied and pure chemistry?
My name is Ann [436]
I hope this helps.
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3 years ago
How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac
Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

,

• using the molar mass of Ca(OH)2 (74g/mol).

,

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

,

• using the molar mass of HNO3 (63.02g/mol).

4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3

So, 7.32g of HNO3 are required.

4 0
1 year ago
A 3.0 g sample of a gas occupies a volume of 1.00L at 100C and 740 torr pressure. The molecular weight of the
SOVA2 [1]

Answer:

94.2 g/mol

Explanation:

Ideal Gases Law can useful to solve this

P . V = n . R . T

We need to make some conversions

740 Torr . 1 atm/ 760 Torr = 0.974 atm

100°C + 273 = 373K

Let's replace the values

0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K

n will determine the number of moles

(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)

n = 0.032 moles

This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?

Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol

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3 years ago
Question 1(Multiple Choice Worth 50 points)
irinina [24]

Answer:

because it's shellfish

Explanation:

A slipper

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3 years ago
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