Answer: The pressure in atmospheres is 0.674 in the container if the temperature remains constant.
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
where,
= initial pressure of gas = 205 kPa
= final pressure of gas = ?
= initial volume of gas = 4.0 L
= final volume of gas = 12000 ml = 12 L (1L=1000ml)
(1kPa=0.0098atm)
Therefore, the pressure in atmospheres is 0.674 in the container if the temperature remains constant.
Answer:
If the question is which can make a buffer, then NH3, NH4Cl should be correct. Because Ammonium (NH4) is conjugate acid of NH3 so they can form an equilibrium which is basically a buffer whose purpose is to resist pH change.
Explanation:
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition =
× 100
% composition =
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
Answer:
Given molecules are vinegar and triglycerides.
Explanation:
The dipole is a vector quantity and it is heading from less electronegative atom to more electronegative atom in a polar covalent bond.
The structures and the bond dipoles in the given molecules are shown below:
<h3>
Answer:</h3>
200 mL
<h3>
Explanation:</h3>
Concept tested: Dilution formula
We are given;
- Concentration of stock solution as 1.00 M
- Volume of the stock solution as 50 mL
- Molarity of the dilute solution as 0.25 M
We are required to calculate the volume of diluted solution;
- The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
- Using the dilution formula we can determine the volume of diluted solution;
M1V1 = M2V2
Rearranging the formula;
V2 = M1V1 ÷ M2
= (1.00 M × 50 mL) ÷ 0.25 M
= 200 mL
Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.