<h2>
Question
</h2>
A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure
<h2>
Answer:</h2>
434.32mL
<h2>
Explanation:</h2>
Using the combined gas law:
= k
Where;
P = Pressure
V = Volume
T = Temperature
k = constant.
It can be deduced that:
= 
= k ---------------------(i)
Where:
P₁ and P₂ are the initial and final pressures of the given gas
V₁ and V₂ are the initial and final volumes of the given gas
T₁ and T₂ are the initial and final temperatures of the gas.
<em>From the question:</em>
the gas is methane
P₁ = 760mmHg
P₂ = 616mmHg
V₁ = 398mL
V₂ = ?
T₁ = 30°C = (30 +273)K = 303K
T₂ = -5°C = (-5 +273)K = 268K
Substitute these values into equation (i) as follows;
= 
Solve for V₂
V₂ = 
V₂ = 434.32mL
Therefore, the volume of the sample at -5C and 616mmHg pressure is 434.32mL
Answer:
A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a simple substitution reaction or simple displacement reaction.
Explanation:
A simple substitution reaction or simple displacement reaction, called single-displacement reaction, is a reaction in which an element of a compound is substituted by another element involved in the reaction. The starting materials are always pure elements and an aqueous compound. And a new pure aqueous compound and a different pure element are generated as products. The general form of a simple substitution reaction is:
AB + C → A +BC
where C and A are pure elements; C replaces A within compound AB to form a new co, placed CB and elementary A.
So, in a Single replacement reaction an uncombined element replaces an element.
<u><em>A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a simple substitution reaction or simple displacement reaction.</em></u>
Answer:
3.3 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 2.7 atm
- Initial volume (V₁): 1.6 L
- Final pressure (P₂): 1.3 atm
Step 2: Calculate the final volume of the balloon
Inside the balloon we have gas. If we consider it behaves as an ideal gas, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 2.7 atm × 1.6 L / 1.3 atm
V₂ = 3.3 L
Because of the plates crashing into each other
Answer:
Answer is in the explanation.
Explanation:
For the reaction:
Ba₃(PO₄)₂(aq) + 3Na₂SO₄(aq) → 3BaSO₄(s) + 2Na₃PO₄(aq)
As Na₂SO₄(aq) is in excess, limiting reactant is Ba₃(PO₄)₂(aq). As the molarity of the solution is 0,25M and you knew the volume of the solution, you can obtain the moles of Ba₃(PO₄)₂ doing 0,25M×volume.
As 1 mol of Ba₃(PO₄)₂(aq) react with 3 moles of BaSO₄ the moles of BaSO₄ are three times moles of Ba₃(PO₄)₂.
As BaSO₄ molar mass is 233,38g/mol. The mass of BaSO₄ is given by moles of BaSO₄ × 233,38g/mol
I hope it helps!
