Answer:
The partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C Is 0.103 atm.
The correct option is A.
Explanation;
NH4I(s) ⇋ NH3(g) + HI(g)Kp = 0.215 at 400°C
NH4I(s)= 0.215
NH3(g)=0.103
HI(g)Kp=0.112
Therefore = 0.103 +0.112= 0.215
Therefore the partial pressure of ammonia at equilibrium is 0.103 atm
So here we are given that the the velocity of the proton ( V ) is 2.0 × 
meters / second, with a magnetic field of strength 5.5 × 
tesla. If they each form a right angle, they are hence perpendicular to one another, such that ....
F = q( V × B ),
F = q v B( sin ∅ ),
F = q v B( sin( 90 ) )
.... they form the following formula. Let's go through each of the variables in our formula here -
{ F = Magnetic Force ( which has to be calculated ), q = charge of proton (has charge of 1.602 × 
coulombs ), B = magnetic field }
All we have to do now is plug and chug,
F = ( 1.602 × )( 2.0 × )( 5.5 × ) = ( About ) 1.8 × 
Newtons
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Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate
