3 years ago

What kind of season would you experience in the northern hemisphere in December

Chemistry

2 answers:

nika2105 [10]3 years ago

5 0

The type of season would be colder than the average temperature for that selected area. It is winter for thar area.

BartSMP [9]3 years ago

4 0

A cold type of season; like winter.

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At what temperature (in °C) would one liter of phosphine gas, PH3, have a mass of 1.725 g, if the gas pressure is 0.9910 atm? As

Nutka1998 [239]

Answer: Temperature of one liter of phosphine gas with a mass of 1.725 g and gas pressure of 0.9910 atm is $-36^0C$

Explanation:

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 0.9910 atm

V= Volume of the gas= 1.0 L

T= Temperature of the gas in kelvin = ?

R= Gas constant = $0.0821Latm/Kmol$

n= moles of gas= $\frac{\text {given mass}}{\text {molar mass}}=\frac{1.725g}{34g/mol}=0.051ol$

$0.9910atm\times 1.0=0.051\times 0.0821\times T$

$T=236.7K=(236.7-273)^0C=-36.3^0C$

Thus the temperature (in °C) of one liter of phosphine gas with a mass of 1.725 g and gas pressure of 0.9910 atm is $-36.3^0C$

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4 years ago

TIME CRUNCH PLS HELP!!!!!When ethane, C2H6 , reacts with chlorine gas the main product is C2H5Cl, but small amounts of C2H4Cl2 a

k0ka [10]

u lost me at time crunch but hey have fun with that :D

7 0

3 years ago

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How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy

malfutka [58]

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

$\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles$

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01 moles of diethyl ether require heat = $\frac{26.5}{1}\times 1.01=26.8kJ$

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

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4 years ago

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k0ka [10]

Answer:

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Explanation:

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3 years ago

What is the name of this molecule?

Komok [63]

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3 years ago

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