The temperature in kelvin does 60.5 liters of sulfur dioxide occupy if there are 2.5 mole at 0.75 atm is 221.07 kelvin
Explanation
This is calculated using ideal gas equation, that is PV=nRT
where, P(pressure) = 0.75 atm
V(volume) = 60.5 L
n(moles) = 2.5 mole
R( gas constant) = 0.0821 L.atm/mol.k
T(temperature =?
by making T the subject of the formula
T is therefore =Pv/nR
T= (0.75 atm x 60.5 L) / ( 2.5 molex 0.0821 L.atm/mol.K) = 221.07 kelvins
Answer:
- Add AgNO₃ solution to both unlabeled flasks: based on solubility rules, you can predict that when you add AgNO₃ to the NaCl solution, you will obtain AgCl precipitate, while no precipitate will be formed from the NaClO₃ solution.
Explanation:
<u>1. Adding AgNO₃ to NaCl solution:</u>
- AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
<u>2. Adding AgNO₃ to NaClO₃ solution</u>
- AgNO₃ (aq) + NaClO₃ (aq) → AgClO₃ (aq) + NaNO₃ (aq)
<u />
<u>3. Relevant solubility rules for the problem.</u>
- Although most salts containing Cl⁻ are soluble, AgCl is a remarkable exception and is insoluble.
- All chlorates are soluble, so AgClO₃ is soluble.
- Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.
Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.
Then, the precipiate will permit to conclude which flask contains AgCl.
It is because they increases linearly due to adding of 1 proton at each step.
Answer:
Explanation:
We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )
Let it be V₂
V₂ = 7.87 litres
22.4 litres of any gas is equivalent to 1 mole
7.87 litres of air will be equivalent to
7.87 / 22.4 moles
= .35 moles .
To find “x” you need to add the 5 to the other side of the equal sign so that x is by itself and so ur answer is 12
