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igomit [66]
3 years ago
6

A stream leaving a mountain range deposits a large part of its load in what

Chemistry
1 answer:
Irina-Kira [14]3 years ago
7 0
Alluvial Fan.  Hope that helps. 
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How many moles are there in 3.612 x 1024 molecules of CaO?
melamori03 [73]

The number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

<h3>Number of moles in the molecules</h3>

The number of moles in  3.612 x 10²⁴ molecules of CaO is calculated as follows;

6.02 x 10²³ molecules = 1 mole

3.612 x 10²⁴ molecules = ?

= (3.612 x 10²⁴ ) / (6.02 x 10²³ )

= 6 moles

Thus, the number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

Learn more about number of moles here: brainly.com/question/15356425

3 0
2 years ago
What would ne the acceleration in a body moving wit uniform velocity and why​
meriva

Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.

Hope this helps :) :)

3 0
3 years ago
Read 2 more answers
A salt forms in the reaction of barium with chlorine what is the most likely formula unit of this salt
Mashcka [7]
 Best Answer:<span>  </span><span>LiCl 

Cl and Li 

glucose 

BeF2 

last question answer 3 strong electrostatic requires a lot of energy
does this help though??</span>
3 0
3 years ago
How many atoms are there in 5.70 mol of hanfnium?
ANTONII [103]

The correct answer would be 3.49 times 10^ minus 24 molecules

6 0
3 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
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