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3 years ago

What effect would a latitude of 60° have on the climate?

Chemistry

1 answer:

Crazy boy [7]3 years ago

6 0

Answer:Temperatures Cool With Increasing Latitude

Explanation:

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1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

We have the following data:

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:

$P_0*V_0 = P*V$

$1*1 = P*0.473$

$1 = 0.473\:P$

$0.473\:P = 1$

$P = \dfrac{1}{0.473}$

$\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

The new pressure of the gas is 2.11 atm

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$\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

3 0

3 years ago

Which will result in positive buoyancy and cause the object to float?

Darina [25.2K]

I would say D. Let me know if i am wrong.

3 0

3 years ago

Read 2 more answers

Which types of reactions would result in the production of gas?

mylen [45]

Answer:

farts

Explanation:

5 0

3 years ago

Which orbital has the lowest energy

GalinKa [24]

Answer:

s orbital

Explanation:

it has the lowest energy because

Paulis law state that orbitals with lower energy must be fill first before that of higher energy

and the s orbital is filled first

5 0

3 years ago

A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o

statuscvo [17]

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = $61.1^{0}C$

Heat absorbed by Scandium = $1.50*10^{3}J$

The equation showing the relationship between heat, mass, specific heat and temperature change:

$Q = m C (deltaT)$

Where Q is heat = $1.50*10^{3}J$

m is mass = 50.0 g

ΔT = $61.1^{0}C$

On plugging in the values and solving for C(specific heat) we get,

$1.50*10^{3}J$ =50.0g(C)( $61.1^{0}C$ )

C = 0.491 $\frac{J}{g^{0}C }$

Specific heat of the metal = 0.491 $\frac{J}{g^{0}C }$

7 0

3 years ago