Answer:
alkene
Explanation:
Since the end only has CH2 it would be an alkene.
Alkene means their is a double bond present.
Carbon always wants 4 bonds (4-2=2) Therefore this shows us that it is a double bond, known as an alkene.
Also,
alcohol = OH
alkane = single bond
alkene = double bond
alkyne = triple bond
- Hope this helps! Please let me know if you want further explanation.
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
Potassium is a chemical element
Symbol: K
Atomic number: 19
Atomic mass: 39.0983 u ± 0.0001 u
Electron configuration: [Ar] 4s1
Melting point: 146.3°F (63.5°C)
Answer:
1. 1.00 gm
2. 50 ml
3. 38.93 ml
4. 11.07 ml
5. 0.01107 L
6. 0.010 moles / L
7. 0.0001107 moles
8. 0.0001107 moles
9. 0.00647042 grams
Explanation:
Silver nitrate can react with various compounds to form different products. The weight of products may be different from the original solution introduced due to combustion reaction, as heat energy is released during the chemical process.