Answer:
Please find the structure attached as an image
Explanation:
Based on the characteristics ending name (-ene) of the organic compound above, it belongs to the ALKENE GROUP. Alkenes are characterized by the possession of a carbon to carbon double bond (C=C) in their structure.
- But-3-ene tells us that the organic compound has four straight carbon atoms with the C=C (double bond) located on the THIRD carbon depending on if we count from right to left or vice versa.
- 2 methyl indicates that the methyl group (-CH3) is located as an attachment on the second carbon (carbon 2).
N.B: In the structure attached below, the counting is from the left to right (→).
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Answer:</h3>
b. 1 Potassium atom
c. 1 bromine atom
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Explanation:</h3>
What is a compound?
- A compound is a substance that is made up of atoms from different elements.
- The formula of a compound shows the elements making up the compound.
- Additionally, the formula of compound shows the number atoms of each element making up the compound.
In this case;
We are given the Formula KBr
- This is the formula of the compound potassium bromide.
- From the formula, we can tell that the compound is made up of potassium and bromine atoms.
- The formula also shows the number of atoms of each element, that is, 1 potassium atom and 1 bromide atom.
Mass of BaO in initial mixture = 3.50g
Explanation:
Let mass of BaO in mixture be x g
mass of MgO in mixture be (6.35 - x) g
Initially CO_2
Volume = 3.50 L
Temp = 303 K
Pressure = 750 torr = 750 / 760 atm
Applying ideal gas equation
PV = nRT
n = PV / RT
(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303
(n)_CO_2 = 0.139 mole
Finally; mole of CO_2
n= PV /RT
((245/760) *3.5) / 303* 0.0821
(n)_CO_2 = 0.045 mole
Mole of CO_2 reacted = 0.139 - 0.045
=0.044 mole
BaO + CO_2 BaCO_3
Mgo + CO_2 MgCO_3
moles of CO_2 reacted = ( moles of BaO + moles of MgO)
moles of BaO in mixture = x / 153 mole
moles of MgO in mixture = 6.35 - x mole / 40
Equating,
x/ 153 +6.35/40 = 0.094
= x/153 + 6.35 / 40 - x/40 =0.094
= x (1/40 - 1153) = (6.35/40 - 0.094)
= x * 10.018464
= 0.06475
mass of BaO in mixture = 3.50g
