Answer:density
Explanation:
it’s how’s how dens the ball is
Answer:
Option A. 1.8×10²⁴ molecules.
Explanation:
Data obtained from the question include:
Number of mole of methane = 3 moles
Number of molecules of methane =?
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ molecules.
Thus, 1 mole of methane equally contains 6.02×10²³ molecules.
With the above information in mind, we can obtain the number of molecules in 3 moles of methane as follow:
1 mole of methane contains 6.02×10²³ molecules.
Therefore, 3 moles of methane will contain = 3 × 6.02×10²³ = 1.8×10²⁴ molecules.
Thus, 3 moles of methane contains 1.8×10²⁴ molecules.
1.4715 atm is the pressure of the sample 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K.
What is an ideal equation?
The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa).
Given data:
Volume (V) = 23.3 L
Number of mole (n) = 1.33 moles
Temperature (T) = 314 K
Gas constant (R) = 0.821 atm.L/Kmol
Pressure (P) =?
The pressure inside the container can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
P × 23.3 L = 1.33 moles × 0.0821 ×314 K
P = 1.4715 atm
Therefore, the pressure of the sample is 1.4715 atm.
Learn more about the ideal gas equation:
brainly.com/question/23826793
SPJ1
Answer:
Graphite and diamond are two of the most interesting minerals. They are identical chemically – both are composed of carbon (C), but physically, they are very different. Minerals that have the same chemistry but different crystal structures are called polymorphs.
Answer:
You need to add 19,5 mmol of acetates
Explanation:
Using the Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [base]/[acid]
For the buffer of acetates:
pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]
As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:
5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]
1,95 = [CH₃COO⁻]/[10]
<em>[CH₃COO⁻] = 19,5 milimoles</em>
Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.
I hope it helps!
