Answer:
CH₅N
Explanation:
In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:
(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C
Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:
(45.0 g)(mol/18.02g) = 2.497...mol H₂O
Moles of H is found using the molar ratio of 2H:1H₂O:
(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H
The ratio of H to C in the compound is:
(4.994...mol H)/(0.99977... mol C) = 5 H:C
Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 18 kg
b) T₁ = 285 K
c) T₂ = 318 K
d) Q = 267.3 kJ
e) S = ?
<u>2) Principles and equations</u>
The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.
The mathematical relation between the specific heat and the heat energy absorbed is:
Where,
- Q is the heat absorbed,
- S is the specific heat, and
- ΔT is the temperature increase (T₂ - T₁)
<u>3) Solution:</u>
<u>a) Substitute the data into the equation:</u>
- 267.3 kJ = 18 kg × S × (318 K - 285 K)
<u>b) Solve for S and compute:</u>
- S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)
The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.
<u>c) Convert to J /( kg . k):</u>
- 0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)
Now we can see that the option A is is the answer, assuming the units.
Answer:
I'm not sure
Explanation:
sorry wish that I did tho
Answer:
K₂CO₃
Explanation:
Given parameters:
Number of moles of K = 0.104mol
Number of moles of C = 0.052mol
Number of moles of O = 0.156mol
Method
From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.
Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.
- Divide the given moles through by the smallest which is C, 0.052mol.
- Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
- This is the empirical formula
Solution
Elements K C O
Number of moles 0.104 0.052 0.156
Dividing by the
smallest 0.104/0.052 0.052/0.052 0.156/0.052
2 1 3
The empirical formula is K₂CO₃
Here we have to calculate the amount of 
ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄ 
g of ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.
