Answer:
w₁ = 0 J
w₂ = -297 J
Explanation:
Step 1: Given data
Initial volume: 266 mL (0.266 L)
Final volume: 983 mL (0.983 L)
Step 2: Calculate the work done (in joules) by the gas if it expands against a vacuum
We will use the following expression.
w₁ = -P₁ × ΔV
Since the gas expands against a vacuum, P₁ = 0. Thus, w₁ = 0 J
Step 3: Calculate the work done (in joules) by the gas if it expands against a constant pressure of 4.09 atm
We will use the following expression.
w₂ = -P₂ × ΔV
w₂ = -4.09 atm × (0.983 L-0.266 L) = -2.93 atm.L
Then, we convert w₂ to Joule using the conversion factor 1 atm.L = 101.325 J.
-2.93 atm.L × 101.325 J/1 atm.L = -297 J
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Answer:
10.051kg of gold is the mass in kilograms
Explanation:
The density of gold is 19.3g/mL. To solve this question we need to convert the volume of the block of gold to grams using its density. Then, we must convert grams to kilograms using the equivalence factor (1000g = 1kg):
<em>Mass in grams:</em>
520.79mL * (19.3g / mL) = 10051g of gold
<em>Mass in kilograms:</em>
10051g of gold * (1kg / 1000g) = 10.051kg of gold
Answer:
H₂O
Explanation:
The empirical formular of the compound is obtained using the following steps;
Step 1: Divide the percentage composition by the atomic mass
Hydrogen = 11.21 / 1 = 11.21
Oxygen = 88.79 / 16 = 5.55
Step 2: Divide by the lowest number
Hydrogen = 11.21 / 5.55 = 2.02 ≈ 2
Oxygen = 5.55 / 5.55 = 1
This means the ratio of the elements is 2 : 1
The empirical formular (simplest formular of a compound) of the compound is;
H₂O
Answer:
Na-O
Be-S
Explanation:
ionic bonds= metal and non metal
