To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.
Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed
Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:
Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s
Answer:
a) ![10 mg NO_3^-/L](https://tex.z-dn.net/?f=10%20mg%20NO_3%5E-%2FL)
b) ![1.61*10^{-4}mol NO_3^-/L](https://tex.z-dn.net/?f=1.61%2A10%5E%7B-4%7Dmol%20NO_3%5E-%2FL)
c) ![2.26 mg N/L](https://tex.z-dn.net/?f=2.26%20mg%20N%2FL)
Carbon: ![C=26.64 \frac{mg C}{L}](https://tex.z-dn.net/?f=C%3D26.64%20%5Cfrac%7Bmg%20C%7D%7BL%7D)
Explanation:
<u>Nitrate</u>
First of all, is important to know that:
![1 ppm=1 mg/L](https://tex.z-dn.net/?f=1%20ppm%3D1%20mg%2FL)
a) 10 ppm of nitrate (
) is equal to ![10 mg NO_3^-/L](https://tex.z-dn.net/?f=10%20mg%20NO_3%5E-%2FL)
b) The molecular weight of nitrate is ![62 g NO_3^-/mol](https://tex.z-dn.net/?f=62%20g%20NO_3%5E-%2Fmol)
![10 mg NO_3^-/L=0.01 g NO_3^-/L](https://tex.z-dn.net/?f=10%20mg%20NO_3%5E-%2FL%3D0.01%20g%20NO_3%5E-%2FL)
![\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%20g%20NO_3%5E-%2FL%7D%7B62%20g%20NO_3%5E-%2Fmol%7D%3D1.61%2A10%5E%7B-4%7D%20mol%20NO_3%5E-%2FL)
c) Nitrate has 14 mg of N per 62 mg of NO3
![10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L](https://tex.z-dn.net/?f=10%20mg%20NO_3%5E-%2FL%2A%5Cfrac%7B14%20mg%20N%7D%7B62%20mg%20NO_3%5E-%7D%3D2.26%20mg%20N%2FL)
<u>Carbon</u>
Carbonate has 12 mg of C per 60 mg of ![CO_3^{-2}](https://tex.z-dn.net/?f=CO_3%5E%7B-2%7D)
Bicarbonate has 12 mg of C per 61 mg of ![HCO_3^{-}](https://tex.z-dn.net/?f=HCO_3%5E%7B-%7D)
![C=24 \frac{mg CO_3^{-2}}{L}*\frac{12 mg C}{60 mg CO_3^{-2}}+111 \frac{mg HCO_3^{-}}{L}*\frac{12 mg C}{61 mg HCO_3^{-}}](https://tex.z-dn.net/?f=C%3D24%20%5Cfrac%7Bmg%20CO_3%5E%7B-2%7D%7D%7BL%7D%2A%5Cfrac%7B12%20mg%20C%7D%7B60%20mg%20CO_3%5E%7B-2%7D%7D%2B111%20%5Cfrac%7Bmg%20HCO_3%5E%7B-%7D%7D%7BL%7D%2A%5Cfrac%7B12%20mg%20C%7D%7B61%20mg%20HCO_3%5E%7B-%7D%7D)
![C=26.64 \frac{mg C}{L}](https://tex.z-dn.net/?f=C%3D26.64%20%5Cfrac%7Bmg%20C%7D%7BL%7D)
Larger nickel–zinc battery systems have been known for over 100 years. Since 2000, development of a stabilized zinc electrode system has made this technology viable and competitive with other commercially available rechargeable battery systems.
Answer:
D
its wings deflect air downward, and the air pushes the plane upward.
Explanation:
According to Newtons third law of motion for every action there is an equal and opposite reaction.
the wings in the plane are designed in such way that the air flowing on the wing is deflected downwards as a result there is a lift produced on the plane due to Newtons third law of motion.
Answer:
A silicon atom has 14 electrons around the nucleus, and of these, there are 4 valence electrons on the outermost orbital. When this is made into a single crystal, it can be used as a material for semiconductor products. When it crystalizes, the nuclei share electrons and they bond with 8 electrons around each nucleus.