Answer:
Substance is Aluminium
Explanation:
We are given;
Mass; m = 13.5 grams
Volume; V = 5 cm³
Formula for density is;
density = m/V
Density = 13.5/5
Density = 2.7 g/cm³
From the table attached, we can see that the element with a corresponding density of 2.7 is Aluminium
Answer:
Buffer B has the highest buffer capacity.
Buffer C has the lowest buffer capacity.
Explanation:
An effective weak acid-conjugate base buffer should have pH equal to 
of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.
Acetic acid is a weak acid and 
is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):
![pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28CH_%7B3%7DCOOH%29%2Blog%5Cfrac%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
Buffer A: 
Buffer B: 
Buffer C: 
So, both buffer A and buffer B has same pH value which is also equal to . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.
The pH of buffer C is far away from . Therefore, buffer C has the lowest buffer capacity.
Wood sinks in ethanol => wood has higher density than ethanol.
wood floats in gasoline=> wood has lower density than gasoline
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
