Answer:
The final pressure of gas is 82.64 KNm⁻²
Explanation:
Given data:
Initial volume of gas = 180 cm³
Temperature of gas = 27°C
Initial pressure = 101 KNm⁻²
Final volume = 220 cm³
Final pressure = ?
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
101 KNm⁻² × 180 cm³ = P₂
× 220 cm³
P₂ = 18180 KNm⁻². cm³/220 cm³
P₂ = 82.64 KNm⁻²
The final pressure of gas is 82.64 KNm⁻².
Explanation:
An oxidation number is a number that is assigned to an atom in a substance. The oxidation number could be positive, negative, or zero, and it indicates if electrons are lost or gained. In other words, the oxidation number is a number that helps us keep track of electrons in an atom.
Answer:
B) 0.59 M NaCl.
Explanation:
- It is known that the no. of millimoles of NaCl before dilution = the no. of millimoles of NaCl after dilution.
∵ (MV) before dilution = (MV) after dilution.
<em>∴ M after dilution = (MV) before dilution / V after dilution </em>= (3.2 M)(25.0 mL)/(135.0 mL) = <em>0.5926 M ≅ 0.59 M.</em>
Answer:
262 ppm of Na₃PO₄
Explanation:
In a dilution, the concentration of the initial solution is decreased. When you take 5.00mL of the solution that is diluted to 25.0mL The solution is diluted 25/5 = <em>5 times</em>
If you make another two serial dilutions the final solution wil decrease its concentration 5*5*5 = 125 times
As original solution containing 0.200 M of Na3PO4, the final solution will have a concentration of:
0.200M / 125 = <em>1.6x10⁻³M</em>
Molarity is defined as the ratio between moles and liters. 1.6x10⁻³ moles of Na3PO4 in 1L are:
1.6x10⁻³mol ₓ (164g/mol) = 0.262g Na₃PO₄ / L
Assuming density of Na3PO4 as 1g/mL the concentration of the solution is:
0.262mL Na₃PO₄ / L
As 1mL = 1000μL:
262μL Na₃PO₄ / L
μL of solute per L of solution is equal to ppm, that means the solution has:
<h3>262 ppm of Na₃PO₄</h3>
