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Kazeer [188]
3 years ago
14

1 What is the approximate altitude of Polaris when viewed from New York City?

Chemistry
1 answer:
77julia77 [94]3 years ago
6 0

Answer:

44° to 45°

Explanation:

The altitude of Polaris star  when viewed from New York City is somewhat between 44° to 45°. However, Polaris is directly overhead at the North Pole (90° of latitude); in other words, the angle between Polaris and the horizon at the North Pole is 90°. This angle is called "the altitude" of Polaris.

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The trend in electronegativity is F >Cl >Br; therefore, the most ionic compound (CH2F2) has the lowest boiling point, and
bulgar [2K]

Answer:

False

Explanation:

Increase in polarity of a molecule leads to higher boiling points. The more polar a molecule is, the higher the energy required to breaks intermolecular forces of attraction hence the higher the boiling point. This is the reason why ionic compounds and compounds having polar covalent bonds in them tend to have high boiling points.

5 0
3 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

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