Answer:
Explanation:
4 = Given data:
Initial volume = 400 mL
Initial pressure = 450 torr
Initial temperature = 210 K
Final temperature = ?
Final volume = 1500 mL
Final pressure = 800 torr
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁/ P₁V₁
T₂ = 800 torr × 1500 mL × 210 K / 450 torr × 400 mL
T₂ = 252000,000 K / 180000
T₂ = 1400 K
5 = Given data:
Initial volume of gas = 4.5 L
Initial temperature = 25°C (25 + 273 = 298 k)
Final temperature = 25°C×3= 75°C (75+273 = 348 k)
Final volume = ?
Solution:
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
V₂ = V₁T₂/T₁
V₂ = 4.5 L × 348 K / 298 k
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
Extra credit:
Given data:
Initial volume = 356 cm³ or 356 mL
Initial pressure = 105000 pa
Initial temperature = 23 °C
Final temperature = ?
Final volume = 560 dL
Final pressure = 36 psi
Formula:
Final volume = 560×100 = 5600 mL
Initial temperature = 23 °C ( 273 + 23 = 296 K)
Final pressure = 36 × 6895 = 248220 Pa
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁/ P₁V₁
T₂ = 248220 Pa × 5600 mL × 296 K /105000 pa × 356 mL
T₂ = 411449472,000 K / 37380000
T₂ = 11007.21 K
Answer:
Homogeneous
Explanation:
Homogeneous mixtures are uniform in composition. They have the same proportion of components throughout. Homogeneous mixtures are called solutions. Sugar, paint, alcohol, gold are all examples of homogeneous mixtures because they look the same throughout.
Answer:
See image attached and explanation
Explanation:
I have attached a detailed mechanism of the reaction to this answer. This reaction occurs by SN1 mechanism. It implies that the transition state involves a carbocation.
However, the initial carbocation formed is a primary carbacation. Remember that the order of stability of carbocations is methyl< primary < secondary< tertiary. This means that tertiary carbocations are the most stable carbocations. Tertiary carbocations are those in which the carbon atom bearing the carbon atom is attached to three other carbon atoms.
In the mechanism below, the substrate converts from a primary to a tertiary cabocation (most stable) by a 1,2-alkyl shift as shown giving the 3-ethoxy-3-methylpentane product.
Answer:
Explanation:
To convert from molecules to moles, we must use Avogadro's Number: 6.022*10²³. This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are molecules of water.
Multiply by the given number of molecules.
Flip the fraction so the molecules of water cancel.
The original measurement of atoms has 2 significant figures ( 4 and 3), so our answer must have the same. For the moles we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.
There are about 7.1 moles of water.