Answer:
80L
Explanation:
V1/T1 = V2/T2
V2 = V1 T2/T1
T1 = 300K
V1 = 60L
T2 = 400K
V2 = ?
V2 = V1 T2/T1
V2 = (60L)(400K) / (300K)
V2 = 80L
Answer:
I don’t know this answer sorry
Explanation:
Answer:
first option is not true
Explanation:
1 mole = 6.02 × 10²³ particles
C3H8 has 1 mole, so has 6.02 × 10²³ particles
5O2 has 5 moles so 5 × 6.02 × 10²³ = 3.01 × 10²⁴ particles
3CO2 has 3 moles so 3 × 6.02 × 10²³ = 1.806 × 10²⁴ particles
4H2O has 4 moles so 4 × 6.02 × 10²³ = 2.408 × 10²⁴ particles
Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.
I believe the answer is C) both
Hope this helps
