Answer:
4.14 x 10²⁴ molecules CO₂
Explanation:
2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O
To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.
4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀
100 grams C₄H₁₀ 1 mol C₄H₁₀ 8 mol CO₂
-------------------------- x ---------------------- x ---------------------
58.124 g 2 mol C₄H₁₀
6.022 x 10²³ molecules
x ------------------------------------ = 4.14 x 10²⁴ molecules CO₂
1 mol CO₂
Answer:
not sure if this is the right way to answer this question but PbS is Phosphate-buffered saline (i think)
Explanation:
Answer:
True
Explanation:
It's true because the pH is a measure of how basic or acid a solution is. In an acidic medium, the pH scales goes from 0 to 7. While in a basic medium goes from 7 to 14. The lower the pH value of the most acid the solution is.
1. The expression pH = -log(molar concentration of hydronium) allow to calculate the pH of a solution.
2. On the other hand, the expression pOH = -log(molar concentration of hydroxide) allow to determine the pOH of a solution.
The values of pH and pOH always obey the following expression:
pH + pOH = 14
Thus if for instance the pH becomes smaller the pOH must become bigger in order to fulfill the equation. Which means that the concentration of hydronium ions is greater than the hydroxide concentration.
For example, in an acidic medium:
if pH= 3, pOH= 11
In this case the molar concentration of hydronium is 0,001M. And the molar concentration of hydroxide ions is just 0,00000000001M.
Answer:
It is given in the question that molarity of the sulphuric acid is 2M, the volume of the sulphuric acid is 20 mL, the volume of the solution is 1`L.
