Answer:
The empirical formula of compound is C₂H₆O.
Explanation:
Given data:
Mass of carbon = 12 g
Mass of hydrogen = 3 g
Mass of oxygen = 8 g
Empirical formula of compound = ?
Solution:
First of all we will calculate the gram atom of each elements.
no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms
no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms
no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms
Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.
C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5
C:H:O = 2 : 6 : 1
The empirical formula of compound will be C₂H₆O
a.
Acids react with bases and give salt and water and the products.
Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
To balance the reaction equation, both sides hould have same number of elements.
Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1
Hence, the reaction equation is already balanced.
b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol
Stoichiometric ratio between NaOH and HCl is 1 : 1
Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol
5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.
Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M
Answer:
a) if the liquid is not vaporized completely, then the condensed vapor in the flask contains the air which is initially occupied before the liquid is heated. When calculating the molar mass of the vapor the moles of air which are initially present are not excluded, so that the molar mass of the vapor would be an increase in value.
b) While weighing the condensed vapor, the flask should be dried. If the weighing flask is not dried then the water which is layered on the surface of the flask is also added to the mass of the vapor. Therefore, the mass of the vapor that is calculated would be increase.
c) When condensing the vapor, the stopper should not be removed from the flask, because the vapor will escape from the flask and a small amount of vapor will condense in the flask. Therefore, the mass of the condensed vapor would be In small value.
d) If all the liquid is vaporized, when the flask is removed before the vapor had reached the temperature of boiling water, then the boiling
temperature of that liquid would be lower than that of the boiling temperature of the water.Therefore, the liquid may have more volatility.
1. Answer;
Equation;
-NaCl(aq) → Na{+}(aq) + Cl{-}(aq)
Explanation;
-The polar property of water allows ionic compounds such as sodium chloride to dissolve in water. (sodium chloride is an ionic compound).
-Sodium chloride dissociates in water into sodium ions and chloride ions as shown by the equation. Water molecules surrounds the negatively charged chloride ions and positive sodium ions and pulls them away into solution.
2. Answer;
Equation;
Fe(ClO4)3(aq) → Fe 3+(aq) + 3ClO4 - (aq)
Explanation;
Iron (iii) perchlorate (Fe(ClO4)3 dissociates in water into iron (iii) ion and perchlorate ion. Iron (iii) ion is positively charged while the perchlorate ion is negatively charged, they are pulled apart by the polar water molecules, a process called dissociation.
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Answer:
a) But-1-ene
b) E-But-2-ene
c) Z-But-2-ene
d) 2-Methylpropene
Explanation:
In this case, if we want to draw the <u>isomers</u>, we have to check the<u> formula </u>
in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain <u>But-1-ene.</u>
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For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain <u>Z-But-2-ene</u>. When the methyls groups are placed on opposite sides we will obtain <u>E-But-2-ene.</u>
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Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain <u>2-Methylpropene.</u>
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See figure 1 to further explanations.
I hope it helps!
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