Answer:
Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g
Explanation:
SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.
the balanced chemical equation is as follows
Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O
Moles of Na₂SO₃ = 
Moles of HCl = 
using mole ratio method to find limiting reagent
For sodium sulfite 
for HCl 
since <u>sodium sulfite</u> is <u>limiting reactant</u> for above chemical reaction
1 mole of Na₂SO₃ produce 1 mole of SO₂
0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂
∴ Mass of SO₂ produce = mole x molar mass of SO₂
= 0.198 x 64
= 12.672 g
Basis of the calculation: 100g
For Carbon:
Mass of carbon = (100 g)(0.80) = 80 g
Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3
For Hydrogen:
Mass of hydrogen = (100 g)(0.20) = 20 g
Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20
Translating the answer to the formula of the substance,
C20/3H20
Dividing the answer,
CH3
The molar mass of the empirical formula is:
12 + 3 = 15 g/mol
Since, the molar mass given for the molecular formula is 30.069 g/mol, the molecular equation is,
C2H6
ANSWER: C2H6
From the electric generator to electric outlets in homes
<em>Answer:</em>
- The concentration of new solution will be 1×10∧-7 M.
<em>Solution:</em>
<em>Data Given </em>
given mass of fluoxymesterone =16.8mg = 0.0168 g
molar mass of fluoxymesterone = 336g/mol
vol. of fluoxymesterone = 500.0 ml = 0.500 L
Stock Molarity of fluoxymesterone = (0.0168/336)÷0.500 = 1×10∧-4 M
So applying dilution formula
Stock Solution : New Solution
M1.V1 = M2.V2
( 1×10∧-4 M) × (1×10∧-6 L) = M2 × 0.001 L
[( 1×10∧-4) × (1×10∧-6)]÷[0.001] = M2
1 × 10∧-7 = M2
<em>Result:</em>
- The concentration of new solution M2 will be 1 × 10∧-7
Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;

If
&





E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol