The rate in m/s is 5.2 * 10^-4 m/s.
<h3>What is the rate in m/s?</h3>
We know that the speed is given as the ratio of the distance covered to the time taken. In this case, we have been told that the rate at which the tide rises is 6.08 ft per hour. We would now need to convert the rate from 6.08 ft per hour to m/s.
Now;
We know that;
1 foot/hour = 8.5 * 10^-5 m/s
6.08 ft per hour = 6.08 ft per hour * 8.5 * 10^-5 m/s/1 foot/hour
= 5.2 * 10^-4 m/s
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Answer:
3 Ca(OH)2 + 2 K3PO4 ---> Ca3(PO4)2 + 6 KOH
Explanation:
Just balance each atom on both sides.
Answer : The number of moles of
produced from the reaction is, 14 moles
Explanation :
The given balanced reaction is,

By the stoichiometry, 1 mole of cesium fluoride react with the 1 mole of xenon hexafluoride to give 1 mole of cesium xenon heptafluoride.
That means the mole ratio of 
As 14 moles of cesium fluoride react with the 14 moles of xenon hexafluoride to give 14 moles of cesium xenon heptafluoride.
Hence, the number of moles of
produced from the reaction is, 14 moles
Total of 127.013 C of charge is passed
Given
weight of Ag solution before current has passed = 1.7854 g
weight of Ag solution after current has passed = 1.8016 g
Molecular mass of Ag = 107.86 g
Faraday's Constant = 96485
First of all we have to apply Faraday's First Law of Electrolysis i.e
m = ZQ
where
Z is propotionality constant (g/C)
Q is charge (C)
Hence,
Z = Atomic mass of substance/ Faraday's Constant
= 
= 0.0011178 g/C
Now ,
change in mass before and after the passing of current (Δm)
Δm = 1.8016g-1.7854g
= 0.0162g
Now amount of coulombs passed = 
amount of coulombs passed = 127.03524 C
Thus from the above conclusion we can say that amount of coulombs have passed is 127.03524 C
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Answer:
The equilibrium will shift to produce less ammonia
Explanation:
According to Le Chatelier's principle, the reaction will try to oppose anything that is done on it, if it was at equilibrium.
When the concentration of H2 is decreased, you are decreasing the concentration of hydrogen so the reaction tries to increase the concentration of hydrogen by breaking down the ammonia on the products side. This will decrease the output of ammonia