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2 years ago

5. Location A is in the center of a large desert. What factor could account for its low annual

1 answer:

4 0

Given what we know about the mechanisms surrounding weather, we can confirm that Large deserts receive less rainfall due to the lack of bodies of water and "rainshadows".

Rainshadow is a term used to describe the effects of nearby mountains on the rainfall amounts of a desert. The mountain causes the air to rise greatly and lose moisture. This moisture is received by the mountain, which means that it does not reach the desert area.

Another limiting factor for the amount of rainfall received by deserts is the lack of bodies of water in the area. Once a desert is formed, it tends to lack large bodies of water. This limits the rainfall it receives given that there is no water nearby to evaporate and cause the formation of clouds.

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Assuming an electron is a hydrogen atom can undergo any of the transitions below, thus changing its energy. Which transition is

inn [45]

Answer:

n = 3 to n = 5

Explanation:

According to the Bohr's model of the atom, electrons in an atom absorb energy to move from a lower to higher energy level.

We must note that as we progress away from the nucleus, the energy levels of electrons become closer together. The energy difference between successive levels decreases and the wavelength of light associated with such transitions become longer.

Hence,the absorption of light of the longest wavelength corresponds to n = 3 to n = 5

5 0

2 years ago

PLZ HELP I WILL GIVE U BRAINLEST PLZ HELP!!!!!!!!!!

sergeinik [125]

My answer is A. I'm probably wrong. in bad in this subject

7 0

2 years ago

Read 2 more answers

Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of

Tpy6a [65]

The volume of O₂ produced: 84.6 L

<h3>Further explanation</h3>

Given

7.93 mol of dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

$\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{3.965\times 0.082\times 321}{1.23365}\\\\V=84.6~L$

5 0

3 years ago

Suppose you need of Grade 70 tow chain, which has a diameter of and weighs , to tow a car. How would you calculate the mass of t

BabaBlast [244]

Answer: check explanation

Explanation:

In this question we are to find mass. In order to calculate the Mass, We need the values of two parameters, that is, the values given for the grade tow chain, and the value given for the mass per length.

Assuming the mass per length is 3 Kilogram per metre(kg/m) and the grade 70 tow chain length is 5 metre(m).

Therefore, the formula for calculating mass of the chain is given below;

Mass of the chain= mass per unit length(kg/m) × length ---------------------------------------------------------------------------------------------------------------------(1).

Mass of the chain= 3 kg/m × 5 m.

Mass of the chain= 15 kg.

7 0

3 years ago

Reaction is

-Dominant- [34]

According to the task, you are proveded with patial pressure of CO2 and graphite, and here is complete solution for the task :
At first you have to find n1 =moles of CO2 and n2 which are moles of C
The you go :
 $CO2(g) + C(s) \ \textless \ =\ \textgreater \ 2CO(g)$

n1 n2 0
-x -x +2x
$n1-x n2-x 2x Kp=P^2 (CO)/Pco2$
After that you have to use the formula $PV=nRT$
$Kp=(2x)^2/n1-x$
Then you have to solve x, and for that you have to use RT/V
And to find total values: $P(total)=n(total)*R*T/V$
I am absolutely sure that this would be helpful for you.

7 0

2 years ago