Answer:
3 CO₂ ......a three-carbon molecule
Explanation:
- Photosynthesis involves the synthesis of food by green plants and algae in the presence of energy from the sun, water, and carbon dioxide.
- The process occurs in two stages known as the light-dependent reactions and light-independent reactions also known as the Calvin cycle.
- During light reactions of photosynthesis, water molecules are split by energy from sunlight to give out oxygen gas and hydrogen ions which are channeled to the Calvin cycle.
- Additionally, ATP and NADPH molecules are released.
- In the Calvin cycle, 3 molecules of carbon dioxide are combined with hydrogen ions to yield three carbon-molecules.
- The process also uses ATP and NADPH from the light reactions.
The volume of 3.40 mol of gas at 33.3 C and 22.2 atm of pressure is 3.85 liter of gas. This problem can be solved by using the PV=nRT or V= nRT/V equation which is the relation between the molar volume, the temperature, and the pressure of gas. In this formula, P is the pressure, R is the universal gas constant ( 0.0821 atm L/mol K), n is the molecule amount, V is the molecular volume, and T is the temperature. The temperature used in this formula must be in Kelvin, therefore we have to convert the Celcius temperature into Kelvin temperature (33.3 C = 306.45 K).
<span>The calculation for the problem above: 3.4*0.0821*306.45/22.2 = 3.85 liter of gas.</span>
At STP 32 g of O₂ would occupy by the same volume as 4 g of He
<h3>Further explanation</h3>
Complete question
At STP 32 g of O₂ would occupy by the same volume as:
- 4.0 g of He
- 8.0 g of CH₄
- 64 g of H₂
- 32 g of SO₂
Standard Conditions
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So the gas will have the same volume if the number of moles is the same
mol of 32 grams of O₂ :
<em>So mol of 4 g He = mol of 32 g O₂</em>
Answer:
9.57 mol.
Explanation:
<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>
<em />
<em>M = (no. of moles of solute)/(V of the solution (L)).</em>
<em></em>
∴ M = (no. of moles of sucrose)/(V of the solution (L)).
1.1 M = (no. of moles of sucrose)/(8.7 L).
<em>∴ no. of moles of sucrose = (1.1 M)(8.7 L) = 9.57 mol.</em>
The mass of cobalt (III) needed is
m = 5.2 L (0.42 mol/L) ( 93 g/mol)
m = 97.65 g
The volume of nitric acid needed is
V = 5.2 L (0.42 mol/L) (3 mol / 1 mol) (1000 mL/1.6 mol)
V = 1968.75 mL
The moles of water produced is
n = 5.2 L (0.42 mol/L) (3 mol / 1 mol)
n = 3.15 moles
