Answer: C
Explanation:
Sunlight
6 CO2 + 6 H2O --------------> C6H12O6 + 6 O2
Chlorophyll
Carbon(IV)oxide Water Glucose Oxygen
Answer:
Empirical formula is CH₂O.
Molecular formula = C₆H₁₂O₆
Explanation:
Given data:
Mass of hydrogen = 3.36 g
Mass of carbon = 20.00 g
Mass of oxygen = 26.64 g
Molar mass of compound = 180.156 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Empirical formula:
It is the simplest formula gives the ratio of atoms of different elements in small whole number
Number of gram atoms of H = 3.36 / 1.01 = 3.3
Number of gram atoms of O = 26.64 / 16 = 1.7
Number of gram atoms of C = 20 / 12 = 1.7
Atomic ratio:
C : H : O
1.7/1.7 : 3.3/1.7 : 1.7/1.7
1 : 2 : 1
C : H : O = 1 : 2 : 1
Empirical formula is CH₂O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = CH₂O = 12×1 + 2× + 16
Empirical formula mass = 30
n = 180.156 / 30
n = 6
Molecular formula = n (empirical formula)
Molecular formula = 6 (CH₂O)
Molecular formula = C₆H₁₂O₆
Answer:
0.1593 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.
P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.
<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>
Answer:
The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.Explanation:
Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂
