Okay so we are given these requirements:
element which can be used to stuff
bottles that enclose ancient paper
must be a gas at room temperature
must be denser than helium
must not react with other elements
The only element that comes into my
mind is:
<span>Argon</span>
Answer:
The answer to your question is 0.4 moles of Oxygen
Explanation:
Data
Octane (C₈H₈)
Oxygen (O₂)
Carbon dioxide (CO₂)
Water (H₂O)
moles of water = ?
moles of Oxygen = 1
Balanced chemical reaction
C₈H₈ +10O₂ ⇒ 8CO₂ + 4H₂O
Reactant Element Products
8 C 8
8 H 8
20 O 20
Use proportions to solve this problem
10 moles of Oxygen ----------------- 4 moles of water
1 mol of Oxygen ------------------ x
x = (4 x 1) / 10
x = 4 / 10
x = 0.4 moles of water
Answer is: there is 2,69·10²³ atoms of bromine.
m(CH₂Br₂) = 39,0 g.
n(CH₂Br₂) = m(CH₂Br₂) ÷ M(CH₂Br₂).
n(CH₂Br₂) = 39 g ÷ 173,83 g/mol.
n(CH₂Br₂) = 0,224 mol.
In one molecule of CH₂Br₂, there is two bromine atoms, so:
n(CH₂Br₂) : n(Br) = 1 : 2.
n(Br) = 0,448 mol.
N(Br) = n(Br) · Na.
N(Br) = 0,448 mol · 6,022·10²³ 1/mol.
n(Br) = 2,69·10²³.
then the electrons and protons would have a even amount of negetive electric charges
Answer:
The two molecules of acetyl-CoA that are produced from a molecule of glucose goes through two turn in the citric acid cycle, one for each molecule of acetyl-CoA.
Explanation:
Glycolysis the process by which a molecule of glucose is broken down in a series of steps to yield two molecules of pyruvate. The overall equation for the reactions of glycolsis is given below:
Glucose + 2NAD+ ----> 2 Pyruvate + 2NADH + 2H⁺
Each of the two pyruvate molecules produced from glucose breakdown is further oxidized to two molecules of acetyl-CoA and CO₂ each.
2 Pyruvate ----> 2 AcetylCoA + 2CO₂
Each of the acetyl-CoA molecule then enters the citric acid cycle for its oxidation. In each turn of the cycle, one acetyl group enters as acetyl-CoA and two molecules of CO₂ leave.
