The function of mordant in the gram staining is to expose grams positive cell to the decolorizer which dissolves the lipids in the cell wall thus allowing the crystals violent-iodine to leach out of the cell. This facilitate the cell subsequently be stained by with safranin.
Cathode rays are a stream of negatively charged particles or electrons in vacuum tubes. They were first seen by Johann Hittorf in 1869. They are named since they are emitted by electrode or cathode. In order to release the electrons, the electrons should first be detached from the atoms of the cathode.
Answer:
11419 J/g/ 11.419 KJ/g
Explanation:
H=MCQ
H=225×2.03×(-15-10)
H=225×2.03(25) Note; negative sign is of no use
H=11419J/g
Answer:
B.) Oxygen is usually -2
Explanation:
Hydrogen is usually +1.
A pure group 1 element is not always +1.
A monoatomic ion can be a range of numbers. However, it must be a charge other than 0.
Answer:
58.9g of SO2 is produced
8g of oxygen remains unconsumed
Explanation:
The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:
CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)
Molar mass of CS2 = 76.139 g/mol
Molar mass of O2 = 15.99 g/mol
Molar mass of SO2 = 64.066 g/mol
Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles
Number of moles of O2 = 30g/15.999 g/mol =1.88 moles
From the chemical reaction
1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2
Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2
Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced
thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2
Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining
Or 8g of oxygen
58.9g of SO2 is produced
oxygen is the limiting
