What is the empirical formula for a compoundwith the molecular formula C6H12Cl2O2?
(1) CHClO (3) C3H6ClO
(2) CH2ClO (4) C6H12Cl2O2
Answers
3. C₃H₆ClO
- Molecular formulas tell you how many atoms of each element are in a compound, while
- Empirical formulas tell you the simplest or most reduced ratio of elements in a compound.
Given: Molecular formula=C₆H₁₂Cl₂O₂
So, the simplest or most reduced ratio for this formula is C₃H₆ClO which can be deduced by dividing the molecular formula by 2 in order to get its empirical formula.
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Answer:
Explanation:
Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.
Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.
Thus the compound with molecular formula will have simplest ratio of atoms as 3:6:1:1 and thus empirical formula will be
.
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Answers
Answer : The pH of a solution is, 9.903
Explanation : Given,
Concentration of hydrogen ion =
pH : It is defined as the negative logarithm of hydrogen ion concentration.
Formula used :
Now put the value of hydrogen ion concentration in this formula, we get the pH of a solution.
Therefore, the pH of a solution is, 9.903
it become
[H+]=0.1
Answers
Answer:
The mass of sodium reacted is 138 grams
Explanation:
Firstly, the chemical reaction should be represented with a balance chemical equation. The chemical equation can be written as follows.
Na = sodium
Cl2 = chlorine gas
Na + Cl2 → NaCl
The balanced equation is
2Na(s) + Cl2(g) → 2NaCl(aq)
Atomic mass of sodium = 23 grams per mol
Atomic mass of chlorine = 35.5 grams per mol
From the chemical equation the molar mass are as follows
Sodium = 2 × 23 = 46 grams
Chlorine = 35.5 × 2 = 71 grams
Sodium chloride = 2 × 23 + 2 × 35.5 = 117 grams
if 46 grams sodium reacted to produce 117 grams of sodium chloride
? grams of sodium react to produce 351 grams sodium chloride
grams of sodium reacted = (351 × 46)/117
grams of sodium reacted = 16146/117
grams of sodium reacted = 138 grams
Answer:
0.138 kg of Na.
Explanation:
Equation of the reaction
2Na(aq) + Cl2(g) --> 2NaCl(s)
By stoichiometry, since 2 moles of Na reacted with 1 mole of Cl2 to produce 2 moles of NaCl. Therefore,
Calculating the number of moles present,
Cl2:
Number of moles = mass/molar mass
Molar mass of Cl2 = 35.5 * 2
= 71 g/mol.
= 213 * 71
= 3 moles.
NaCl:
Molar mass = 23 + 35.5
= 58.5 g/mol.
Number of moles = mass/molar mass
= 351/58.5
= 6 moles.
Since they both have equal moles,
Mass of Na = number of moles * molar mass
Molar mass of Na = 23 g/mol
= 23 * 6 moles
= 138 g
= 0.138 kg of Na.
Answers
hope this helps!
Answers
If you have the coordinates of the 2 points, then the distance between them is
√ of [ (the difference in their x-values)² + (the difference in their y-values)² ]
A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure.
How many grams of Ny must react to form 1.7 grams of ammonia, NH3?
0.052 g
1.49
0.00589
2.8 g
Answers
Answer:
Explanation:
Here we have to use stoichiometry.
First of all, we have to calculate the mass of 100% of yield:
1.7 g ------- 98%
X -------- 100%
X = 1.73 g (approximately)
Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.
Molar Mass N2 : 14x2 = 28 g/mol
Molar Mass NH3: 14 + 3 = 17 g/mol
28g (N2) ------- 17x2 (NH3)
X ------------ 1.73 g
X = 1.42 g (approximately)