Answer:
n = 0.3 mol
Explanation:
Given data:
Volume of gas = 8.0 L
Temperature of gas = 45 °C (45+273 = 318 K)
Pressure of gas = 0.966 atm
Moles of gas present = ?
Ideal gas constant = R = 0.021 atm.L/mol.K
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Now we will put the values:
0.966 atm × 8 L = n × 0.0821 atm.L/mol.K × 318 K
7.728 atm.L = n × 26.12 atm.L/mol
7.728 atm.L / 26.12 atm.L/mol = n
n = 0.3 mol
Answer:
= 3.78 g H₂O
Explanation:
2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O
2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane
3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen
Limiting Reactant:
A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.
moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035
moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038
Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.
From the equation stoichiometry ...
2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O
then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole
Converting to grams of water produced
= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O
Answer:
I thinks is B
Explanation:
they just look harmful
I also researched and it says its harmful and causes sickness and death
Answer:
C₆H₁₂O₆ and O₂
Explanation:
To know which option is correct, it important that we know the concepts of chemical equation .
A chemical equation gives us an expression of a chemical reaction in terms of symbols and formula. There are two sides to every equation. The side bearing the reactant and the side bearing the product.
The reactant is located on the left side and the product is located on the right side.
Reactant —> Product
With the above information, let us answer the question given above.
6CO₂ + 6H₂O —> C₆H₁₂O₆ + 6O₂
The products are located on the right side of the equation.
Thus, the products of the reaction is C₆H₁₂O₆ and O₂
