Answer:
This solution acts as an efficient buffer
Explanation:
the pH of a buffer solution can be described like this: ![pH=pKa+log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
[acid]=[acetic acid]=
[base]=[sodium acetate]=
replacing, 
If we add an acid, pH will decrease a little bit and if we add a base, pH wil increase a little bit.
lets supose that we change the rate by increasing [base] to 0.1, then
and now lets supose that we increase [acid] to 0.1 
Big changes in concentration of base or acid doesn´t produce big changes in pH, in that way the mix of sodium acetate with acetic acid is a good buffer solution.
Answer:
0.416 mol CaBr₂
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
83.1 g CaBr₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar mass of Br - 79.90 g/mol
Molar Mass of CaBr₂ - 40.08 + 2(79.90) = 199.88 g/mol
<u>Step 3: Convert</u>
<u />
= 0.415749 mol CaBr₂
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.415749 mol CaBr₂ ≈ 0.416 mol CaBr₂
Since you have not included the chemical reaction I will explain you in detail.
1) To determine the limiting agent you need two things:
- the balanced chemical equation
- the amount of every reactant involved as per the chemical equation
2) The work is:
- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.
- determine the number of moles of each reactant with this formula:
number of moles = (mass in grams) / (molar mass)
- set the proportion with the two ratios (theoretical moles and actual moles)
- compare which reactant is below than the stated by the theoretical ratio.
3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:
i) Chemical equation: H₂ + O₂ → H₂O
ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O
iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂
iv) Covert 100 g of H₂ into number of moles
n = 100g / 2g/mol = 50 mol of H₂
v) Convert 100 g of O₂ to moles:
n = 100 g / 32 g/mol = 3.125 mol
vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂
vii) Compare the two ratios:
2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂
Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.
Therefore, the O₂ is the limiting reactant in this example.
Answer:
3.7mL is the volume of the object
Explanation:
To convert the mass of any object to volume we must use density that is defined as the ratio between mass of the object and the space that is occupying. For an object that weighs 7.9g and the density is 2.28g/mL, the volume is:
7.9g * (1mL / 2.28mL) =
<h3>3.7mL is the volume of the object</h3>
