Answer:
8.22E⁰ kg
Explanation:
Given data:
Mass of glucose = ?
Number of molecules = 2.75×10²⁵ molecules
Solution:
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.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For 2.75×10²⁵ molecules of glucose:
One mole of glucose = 180 g
180 g/6.022 × 10²³ molecules× 2.75×10²⁵ molecules
82.2 × 10² g
8220 g
8.22 Kg
8.22E⁰ kg
The balanced chemical equation of the reaction described above is,
C2H6O + O2 --> H2O + C2H4O2
If we have 3.84 g of oxygen, we divide by its molar mass.
n = (3.54 g Oxygen gas) x (1 mole O2/ 32 g O2)
n = 0.11 moles O2
Using ratio and proportion,
number of moles of ethanol = (0.11 moles O2) x (1 mole C2H6)
= 0.11 moles C2H6
Then, we multiply the calculated value to its molar mass, 46 grams /mol.
mass of ethanol = (0.11 mol) x (46 grams / mol)
= <em>5.06 grams</em>
Answer:
They are synthesized from raw materials
Explanation:
All synthesized products goes through lots of processes. This processes involves the conversion of substances known as raw materials into a more useful and modified form.
These synthesized products are usually more useful and made in order to solve a given problem or offer a particular service. Raw materials are cheaper than the synthesized products and are common in all industries and fields.
Answer:
Amount of heat released when 25.0 g of (rubbing alcohol) is combusted equal to 98.7 kJ
Explanation:
According to balanced equation (as given in problem), combustion of 2 moles of rubbing alcohol release 474.28 kJ of heat.
Molar mass of (rubbing alcohol) = 60.095 g/mol
We know, number of moles of a compound is the ratio of mass to molar mass of that compound.
So, 2 moles of (rubbing alcohol) = (rubbing alcohol) = 120.19 g of (rubbing alcohol)
So, combustion of 120.19 g of rubbing alcohol release 474.28 kJ of heat.
Hence amount of heat released when 25.0 g of (rubbing alcohol) is combusted =
Answer:
The answer to your question is 9.075 g of CO₂
Explanation:
Data
mass of C₃H₈ = 39 g
mass of O₂ = 11 g
Balanced chemical reaction
C₃H₈ + 5O₂ ⇒ 3CO₂ + 4H₂
-Calculate the molar mass of the reactants
C₃H₈ = (12 x 3) + (8 x 1) = 36 + 8 = 44 g
O₂ = (16 x 2) = 32 g
-Calculate the limiting reactant
theoretical yield C₃H₈ / O₂ = 44/5(32) = 44/ 160 = 0.275
experimental yield C₃H₈/O₂ = 39/11 = 3.5
From the previous result, we conclude that the limiting reactant is O₂ because the experimental yield was higher than the theoretical yield.
-Calculate the mass of CO₂
160 g of O₂ ----------------- 3(44) g of CO₂
11 g of O₂ ------------------ x
x = (11 x 3(44)) / 160
x = 1452 / 160
x = 9.075 g of CO₂