Answer:
P₂ = 28.5 torr
Explanation:
Given data:
Initial pressure = 38 torr
Initial volume = 500 L
Final volume = 677 L
Final pressure = ?
Solution:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = Initial volume
P₂ = Final pressure
V₂ = Final volume
Now we will put the vales in formula.
P₁V₁ = P₂V₂
P₂ = P₁V₁ /V₂
P₂ = 38 torr × 500 L / 667 L
P₂ = 19000 torr. L / 667 L
P₂ = 28.5 torr
Answer:
Total cost = $40.25 (Approx)
Explanation:
Given:
Per ounce = $1.15
Number of student = 28
Each student eat = 0.035 kg
Find:
Total cost
Computation:
Total weight of candy = 28 × 0.035 kg
Total weight of candy = 0.98 kg
1 ounce = 0.028 kg (approx).
Total weight of candy = 0.98 kg / 0.028
Total weight of candy = 35 ounce (Approx)
Total cost = 35 × $1.15
Total cost = $40.25 (Approx)
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
Explanation:
Compressors and limiters are used to reduce dynamic range — the span between the softest and loudest sounds. Using compression can make your tracks sound more polished by controlling maximum levels and maintaining higher average loudness.