Yes there is by the way are there any answers to choose from
Answer:
1.2 atm
Explanation:
Given data
- Volume of the gas in the tank (V₁): 200.0 L
- Pressure of ethylene gas in the tank (P₁): ?
- Volume of the gas in the torch (V₂): 300 L
- Pressure of the gas in the torch (P₂): 0.8 atm
If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.

Answer:
This question appears incomplete
Explanation:
However, it should be noted that addition of soluble salts generally lowers the freezing point of water hence after the addition, water will no longer freeze at 0°C but lower.
Soluble salts tend to form more ions in water, it is these ions that are responsible for interfering with the hydrogen bonds hence lowering the freezing. Thus, (since each bag are of the same weight) <u>the bag that contains the salt that ionizes more in water will lower the freezing point by the greatest amount</u>.
NOTE: Different weight of the salts could lead to more ions been formed in the water by some salts as against the other.
The number of moles present in 29.5 grams of argon is 0.74 mole.
The atomic mass of argon is given as;
Ar = 39.95 g/mole
The number of moles present in 29.5 grams of argon is calculated as follows;
39.95 g ------------------------------- 1 mole
29.5 g ------------------------------ ?

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.
<em>"Your question seems to be missing the correct symbol for the element" </em>
Argon = Ar
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Data Given:
Time = t = ?
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 107.86/1 = 107.86 g
Amount Deposited = W = 17.3 g
Solution:
According to Faraday's Law,
W = I t e / F
Solving for t,
t = W F / I e
Putting values,
t = (17.3 g × 96500) ÷ (10 A × 107.86 g)
t = 1547.79 s
t = 1.54 × 10³ s