The balanced chemical reaction is written as:
4Al + 3O2 = 2Al2O3
To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:
moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2
Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.
Hybridization refers to the mixing of atomic orbitals in an atom. The number of hybrid orbitals needs to be equal to the number of orbitals that have involved in prior to mixing.
The isolated atoms cannot prevail in a hybridized state as the atom in an isolated state do not form any kind of bond with the other atom, due to which the atomic orbitals do not go through the process of hybridization.
Answer : The correct option is, (A)
because one of each is produced every time an
transfers from one water molecule to another.
Explanation :
As we know that, when the two water molecule combine to produced hydronium ion and hydroxide ion.
The balanced reaction will be:

Acid : It is a substance that donates hydrogen ion when dissolved in water.
Base : It is a substance that accepts hydrogen ion when dissolved in water.
From this we conclude that, the hydrogen ion are transferred from one water molecule to the another water molecule to form hydronium ion and hydroxide ion. In this reaction, one water molecule will act as a base and another water molecule will act as an acid.
Hence, the correct option is, (A)
Answer:
5' RG GWCCY 3'
3' YCCWG GR 5'
Explanation:
The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).
The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'
target Sequence: 5' RGGWCCY 3'
3' YCCWGGR 5'
The enzyme cleavage point is at:
5' RG^GWCCY 3'
3' YCCWG^GR 5'
The product of the cleavage will give a sticky end Cleavage:
5' RG GWCCY 3'
3' YCCWG GR 5'
Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.
Answer:
Molecular formula → PbSO₄ → Lead sulfate
Option c.
Explanation:
The % percent composition indicates that in 100 g of compound we have:
68.3 g of Pb, 10.6 g of S and (100 - 68.3 - 10.6) = 21.1 g of O
We divide each element by the molar mass:
68.3 g Pb / 207.2 g/mol = 0.329 moles Pb
10.6 g S / 32.06 g/mol = 0.331 moles S
21.1 g O / 16 g/mol = 1.32 moles O
We divide each mol by the lowest value to determine, the molecular formula
0.329 / 0.329 = 1 Pb
0.331 / 0.329 = 1 S
1.32 / 0.329 = 4 O
Molecular formula → PbSO₄ → Lead sulfate