Answer:
1.84 L
Explanation:
Using the equation for reversible work:
![W = -P*(V_{2} - V_{1})](https://tex.z-dn.net/?f=W%20%3D%20-P%2A%28V_%7B2%7D%20-%20V_%7B1%7D%29)
Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*(
- 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
Answer:
the work input is depented on the work output
Explanation:
Answer:
<h3>Theanswer is 6 moles</h3>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
![n = \frac{N}{L} \\](https://tex.z-dn.net/?f=n%20%3D%20%20%5Cfrac%7BN%7D%7BL%7D%20%20%5C%5C%20)
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
![n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\](https://tex.z-dn.net/?f=n%20%3D%20%20%5Cfrac%7B3.612%20%5Ctimes%20%20%7B10%7D%5E%7B24%7D%20%7D%7B6.02%20%5Ctimes%20%20%7B10%7D%5E%7B23%7D%20%7D%20%20%5C%5C%20)
We have the final answer as
<h3>6 moles</h3>
Hope this helps you
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide