Answer:
Distance cover by school bus = 149.5 miles
Explanation:
Given:
Velocity of school bus = 65 mph
Time taken by school bus = 2.3 hours
Find:
Distance cover by school bus
Computation:
Distance cover = Velocity x Time taken
Distance cover by school bus = Velocity of school bus x Time taken by school bus
Distance cover by school bus = 65 x 2.3
Distance cover by school bus = 149.5 miles
Answer:
Option A. 9.4 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 8 L
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 343 K
Final volume (V₂) =?
V₁ / T₁ = V₂ / T₂
8 / 293 = V₂ / 343
Cross multiply
293 × V₂ = 8 × 343
293 × V₂ = 2744
Divide both side by 293
V₂ = 2744 / 293
V₂ = 9.4 L
Therefore, the final volume of the gas is 9.4 L
Answer:
Explanation:
To solve this problem, we can use the Combined Gas Laws:
Data:
p₁ = 1.7 kPa; V₁ = 7.5 m³; T₁ = -10 °C
p₂ = ?; V₂ = 3.8 m³; T₂ = 200 K
Calculations:
(a) Convert temperature to kelvins
T₁ = (-10 + 273.15) K = 263.15 K
(b) Calculate the pressure
Atoms
Explanation:
Chemical bonds results from the rearrangement of atoms in a chemical species.
It deals with the various attractive forces joining chemical species togethe.
- When atoms are re-arranged, they form chemical bonds that leads to production of new compounds.
- This is made possible by the exchange or sharing of electrons.
- The driving force for most interatomic bonding is the tendency to have completely filled outer energy levels like the noble gases.
- When atoms are re-arranged in compounds they lead to the production of chemical bonds.
learn more:
Ionic bonds brainly.com/question/6071838
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Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of butane
b) moles of oxygen
According to stoichiometry :
2 moles of butane require 13 moles of 
Thus 0.09 moles of butane will require =
of
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
