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suter [353]
3 years ago
14

Which part of the water cycle would follow step C in the diagram shown?

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

Evaporation

Explanation:

From point C to D, this is a phase change from liquid to gas and it would be depicted by evaporation.

The water cycle presents the transformation of water from one form to another on our planet.

  • From Point A to B in which water falls from clouds is a precipitation step.
  • Point B to Point C is the step in which surface water runs into lakes.
  • Point C to Point D where water turns to gas in the atmosphere is evaporation.
  • The from point a D back to A is condensation.

Learn more:

Evaporation brainly.com/question/10972073

#learnwithBrainly

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Changes in the forms of energy are called
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The answer is B ( you could have just looked it up online)
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Write the complete/total ionic equation for the precipitation reaction that occurs when aqueous iron(II)nitrate is added to aque
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<span>The equation that describes the problem is  Fe(NO3)3(aq) + 3NaOH(aq) ---> Fe(OH)3(s) +  3 NaNO3(aq)
 
The Net ionic equation is written as follows:
 Fe^3(aq) + 3NO3-(aq) + 3Na+(aq) + 3OH-(aq) ---> Fe(OH)3(s) + 3Na+(aq) + 3NO^3-(aq)</span>
6 0
3 years ago
A gas has a volume of 50.0 cm3 at a temperature of -73°c. what volume would the gas occupy at a temperature of -123°c if the pre
vaieri [72.5K]

Boyle’s Law illustrates the inverse relationship of volume and pressure. It follows the formula  p1V1 = P2V2 , where P1V1 denotes initial pressure and volume  and P2V2 denotes values of pressure and  volume.

Now, let us work out for what is asked above.
a. if the pressure is doubled
50.0 p = V x 2p 
V = 50.0 p / 2p

= 50.0 /2

= 25.0 m^3 
b. if the pressure is cut in half
50.0 p = V x p/2 
100 p = V x p 
V = 100 m^3 
c. if the pressure is tripled
50.0 p = V x 3p 
V = 50.0 p / 3p

= 50.0 /3

=16.7 m^3 

<span> </span>

7 0
3 years ago
Write a nuclear equation for the beta decay of Promethium-165
padilas [110]

Answer:

see your answer in pic

Explanation:

4 0
3 years ago
What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
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