Answer:
V₂ = 46.34 mL
Explanation:
Given data:
Initial temperature = 273 K
Initial volume = 50.0 mL
Final temperature = 253 K
Final volume = ?
Solution:
The given problem will be solve through the Charles Law.
Charles 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
by putting values in formula,
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 50.0 mL × 253 K / 273 k
V₂ = 12650 mL.K / 273 K
V₂ = 46.34mL
Answer:
Calculate the volume (in mL) of the 1.356 M stock NaOH solution needed to prepare 250.0 mL ... Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar ... g of glucose in enough water to make 500.0 mL of solution. • Step 2: Transfer 18.6 mL of glucose
Answer:
2MnO2 (s)
Explanation:
3KCN(aq) + 2KMnO4(aq) + 1H2O(l) → 3KCNO(aq) + 2MnO2(s) + 2KOH(aq)
Reactants:
K: 5
C: 3
N: 3
Mn: 2
O: 9
H: 2
Products:
K: 5
C: 3
N: 3
Mn: 2
O: 9
H: 2
Answer:
C
B
D
A
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