Answer:
A
Explanation:
In all the other options either Carbon has more than 4 atoms attached or less than 4.
Carbon can not form more than four bonds. It can only share 4 electrons .
In other option C, No H-atom is linked in the right most Carbon.
whereas according to the definition of Hydrocarbon it should have been a Hydrogen atom.
Many atoms if they are radioactive isotopes will loss protons and neutrons as radiation in order to gain and become more stable
hope that helps
Answer:
The answer is 6.25g.
Explanation:
First create your balanced equation. This will give you the stoich ratios needed to answer the question:
2C8H18 + 25O2 → 16CO2 + 18H2O
Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:
7.72 g / 16 g/mol = 0.482 mol
Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:
x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2
x = 0.347 mol H2O
The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:
0.347 mol x 18 g/mol = 6.25g
Answer:
619°C
Explanation:
Given data:
Initial volume of gas = 736 mL
Initial temperature = 15.0°C
Final volume of gas = 2.28 L
Final temperature = ?
Solution:
Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)
Initial temperature = 15.0°C (15+273 = 288 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = T₁V₂/V₁
T₂ = 2.28 L × 288 K / 0.736 L
T₂ = 656.6 L.K / 0.736 L
T₂ = 892.2 K
K to °C:
892.2 - 273.15 = 619°C
Answer:
No
Explanation:
The Earth would not have seasons if there is no revolution because the temperatures would not change.
