<span>Ionic bonding between sodium and phosphate ions.</span>
In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.
Explanation:
In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.
I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.
It can be said that only 38.3% of energy is put in ATP molecules.
Answer:
102g of crystals
Explanation:
When the Cr(NO₃)₃⋅9H₂O is dissolved in water at 15°C, the maximum mass that water will dissolve in the equilibrium is 208 g per 100g of water. When you heat the water, this mass will increases.
In this problem, at 35°C the water dissolves 310g in 100g of water, as in the equilibrium at 15°C the maximum mass is 208g, the mass of crystals that will form is:
310g - 208g = <em>102g of crystals</em>
<em>-Crystals are the Cr(NO₃)₃⋅9H₂O that is not dissolved-.</em>
I hope it helps!
Molar mass:
HF = 1 + 19 = 20.0 g/mol
Number of moles :
124 / 20.0 => 6.2 moles
Volume = 2.4 L
M = n / V
M = 6.2 / 2.4
M = 2.6 M
Answer A
hope this helps!
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant
3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO
From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂
(b) Mass of NO reacted
(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO
