The Correct Answer is Eletric Current I think
Answer:
True
Explanation:
Dispensing chemical from large stock bottles into smaller container makes sure that students only takes the quantity they need which reduces wastage. Also in situation where the students did not finish the reagent or chemical it is easier for student to pour back into smaller bottle than the entire reagent bottle which can contaminate the whole solution. Small bottles prevent or reduces the risk of spillage because it is easier to handle and pour.
<span>air = 70 % N2
exhaled air =70%N2 + some less O2 compared to only air
2H2O2=2H2O+O2
NaHCO3=2NaHCO3 → Na2CO3 + H2O + CO2 = No O2 gas from decomposition
ssoooo NaHCO3 <exhaled air <air <H2O2
SO sorry for the long wait, hope I helped.
</span>
Answer:
The correct answer is - (a) isobutene.
Explanation:
A 1° or primary hydrogen atom is one that is bonded to a 1° carbon atom; a 2° hydrogen atom is one that is connected to a 2° carbon atom; In isobutene only among all the all option having only primary hydrogen atoms. rest have secondary or teritary hydrogen atoms in them
C1∘H3≡CH: (only 1∘ hydrogen atoms)
<h3>
Answer:</h3>
Al₂(SO₄)₃
<h3>
Explanation:</h3>
We are given percentage composition of elements in a compound;
- Aluminium is 15.77%
- Sulfur is 28.11 %
- Oxygen is 56.12%
We are required to calculate the empirical formula of the compound.
- Assuming the mass of the compound is 100 g then the masses of the elements is;
Aluminium = 15.77 g
Sulfur = 28.11
Oxygen = 56.12
We can determine the number of moles of each;
Moles of Aluminium = 15.77 g ÷ 26.98 g/mol
= 0.585 moles
Moles of sulfur = 28.11 g ÷ 32.07 g/mol
= 0.877 moles
Moles of Oxygen = 56.12 g ÷ 16.0 g/mol
= 3.5075 moles
- But, the empirical formula is the simplest whole number ratio of elements in a compound.
- Therefore; we need to get the ratio of moles of the above elements;
Aluminium : Sulfur : Oxygen
0.585 mol : 0.877 mol : 3.5075 mol
0.585/0.585 : 0.877/0.585 : 3.5075/0.585
1 : 1.5 : 6
But, we need whole number ratios, therefore;
= (1 : 1.5 : 6 ) × 2
= 2 : 3 : 12
Therefore; the formula of the compound is Al₂S₃O₁₂
The compound is written as Al₂(SO₄)₃
Thus, the empirical formula of the compound is Al₂(SO₄)₃
