Answer:
C₁₁H₁₂NO₄
Explanation:
In order to determine the empirical formula of doxycycline, we need to follow a series of steps.
Step 1: Determine the centesimal composition
C: 59.5 mg/100 mg × 100% = 59.5%
H: 5.40 mg/100 mg × 100% = 5.40%
N: 6.30 mg/100 mg × 100% = 6.30%
O: 28.8 mg/100 mg × 100% = 28.8%
Step 2: Divide each percentage by the atomic mass of the element
C: 59.5 /12.0 = 4.96
H: 5.40/1.00 = 5.40
N: 6.30/14.0 = 0.450
O: 28.8/16.0 = 1.80
Step 3: Divide all the numbers by the smallest one
C: 4.96/0.450 = 11
H: 5.40/0.450 = 12
N: 0.450/0.450 = 1
O: 1.80/0.450 = 4
The empirical formula of doxycycline is C₁₁H₁₂NO₄
-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.
Explanation:
Given:
mass of Na2O2 = 25 grams
atomic mass of Na2O2 = 78 gram/mole
number of mole = 
= 
=0. 32 moles
The balanced equation for the reaction:
2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ
It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.
similarly 0.3 moles of Na2O2 on reaction would give:
= 
x = 
= -20.16 KJ
Thus, - 20.16 KJ of energy will be released.
1) Chemical reaction 1: 4Cu + O₂ → 2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ → 2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.
Answer:
The equilibrium between the two forms of the gas is disturbed at high temperatures.
Answer:
Explanation:
The pressure at the bottom of the tank is:
The force exerted on the circular bottom is:
![F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}](https://tex.z-dn.net/?f=F%3D%2873581.921%5C%2CPa%29%5Ccdot%20%28%5Cfrac%7B%5Cpi%7D%7B4%7D%20%29%5Ccdot%20%5B%2812%5C%2Cft%29%5Ccdot%20%28%5Cfrac%7B0.305%5C%2Cm%7D%7B1%5C%2Cft%7D%20%29%5D%5E%7B2%7D)
