Answer:
43.2 moles of carbon dioxide are required and 421g of glucose could be produced
Explanation:
Based on the reaction:
6CO2 + 6H2O → C6H12O6 + 6O2
1 mole of glucose, C6H12O6, requires 6 moles of carbon dioxide. 7.2moles of glucose requires:
7.2mol C6H12O6 * (6mol CO2 / 1mol C6H12O6) =
<h3>43.2 moles of carbon dioxide are required</h3><h3 />
618g of CO2 -Molar mass: 44.01g/mol- are:
618g * (1mol / 44.01g) = 14.04moles CO2
Moles C6H12O6:
14.04moles CO2 * (1mol C6H12O6 / 6mol CO2) = 2.34moles C6H12O6
Mass glucose -Molar mass: 180.156g/mol-
2.34moles C6H12O6 * (180.156g / mol) =
<h3>421g of glucose could be produced</h3>
A benefit would be there would be more money for exploration and a con is that there would be a language barrier between the astronauts
Answer:
Mass = 547.02 × 10⁻²³g
Explanation:
Given data:
Number of atoms of Al = 122 atom
Mass in gram = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
122 atom/6.022 × 10²³ atoms × 1 mol
20.26× 10⁻²³ mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 20.26× 10⁻²³ mol × 27 g/mol
Mass = 547.02 × 10⁻²³g
Answer:
The stronger acid is HBrO3.
Explanation:
It has an additional oxygen making it more electronegative, in turn making it a stronger acid.
An aqueous solution of potassium sulfate exhibits colligative properties. Colligative properties are properties that depends on the concentration of a substance in a solution. These properties are freezing point depression, vapor pressure lowering, osmotic pressure and boiling point elevation. For this problem we use the concept of freezing point depression since we are given the freezing point of the solution. Freezing point depression is as:
ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
<span>m = 0.4014
Thus, the molality of the solution is 0.4014.</span>
