Answer:
Foam fight is a chemical change.
Explanation:
Please give me brainliest :)
False
Although we use many of their ideas to describe atoms today, such as the existence of a tiny, dense nucleus in an atom (proposed by Rutherford), or the notion that all atoms of an element are identical (proposed by Dalton), some of their ideas have been rejected by the modern theory of the atom.
For example, Thompson came up with the plum pudding model to describe an atom, which resembled a sphere of positive charge with electrons embedded in it. We know now, however, that atoms are mostly empty space with a tiny, dense nucleus.
Another example is Dalton's atomic theory, which stated that atoms are indivisible particles. However, this was disproved by the discovery of subatomic particles.
Answer:
Molecular formula is : C₆H₈O₆
Explanation:
40.9 % of C, 4.6 % of H, 54.5 % of O means that in 100 g of ascorbic acid, we have 40.9 g of C, 4.6 g of H and 54.5 g of O.
If 1 mol of ascorbic acid weighs 176 g, let's determine the rules of three to know the grams of each element:
100 g of compound has ___ 40.9 g C __ 4.6 g H ___ 54.5 g O
176 g compound mus have:
(176 . 40.9) / 100 = 72 g of C
(176 . 4.6) / 100 = 8 g of H
(176 . 54.5) / 100 = 96 g of O
If we convert the mass to moles:
72 g . 1 mol / 12 g = 6 C
8 g . 1 mol/ 1g = 8 H
96 g . 1mol / 16 g = 6 O
Molecular formula is : C₆H₈O₆
In conclussion, 1 mol of ascorbic acid has 6 moles of C, 8 moles of H and 6 moles of O
Answer:
2.893 x 10⁻³ mol NaOH
[HCOOH] = 0.5786 mol/L
Explanation:
The balanced reaction equation is:
HCOOH + NaOH ⇒ NaHCOO + H₂O
At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.
The moles of base added is calculated as follows:
n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH
Extra significant figures are kept to avoid round-off errors.
Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.
(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH
The concentration of HCOOH to the correct number of significant figures is then calculated as follows:
C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L
The question also asks to calculate the moles of base, so we convert millimoles to moles:
(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH