Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
Answer:
20.4 grams Zn
Explanation:
To find the mass, you first need to find the moles. This can be found using the Ideal Gas Law equation:
PV = nRT
In this equation,
-----> P = pressure (atm)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)
-----> T = temperature (K)
Before you can plug the values into the equation, you need to convert Celsius to Kelvin.
P = 0.980 atm R = 0.08206 atm*L/mol*K
V = 7.80 L T = 25.0 °C + 273.15 = 298.15 K
n = ? moles
PV = nRT
(0.980 atm)(7.80 L) = n(0.08206 atm*L/mol*K)(298.15 K)
7.644 = n(24.466)
0.312 moles = n
Now that you have the number of moles, you can convert it to grams using the atomic mass of zinc. The final answer should have 3 sig figs to match the sig figs in the given values.
Atomic Mass (Zn): 65.380 g/mol
0.312 moles Zn 65.380 grams
------------------------- x ------------------------- = 20.4 grams Zn
1 mole
TITRATION is the process of reaching equilibrium between acids and bases.
As in math or
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The class of compounds that ammonia belongs to is it is categorized as a basic compound. As it posses a pH value or rating above 7, which is characteristic of basic substances, solutions.
