Answer:
the atom
Explanation:
The bread and cheese separately represent the atom because both bread and cheese are different from one another and we cannot assume it as a molecule because molecules formed when the group of atoms combine together by making bonds with each other and we know that bread and cheese did not make bonds with each other, they only attached, so we called them atoms not molecule.
Answer:
Salt water
Explanation:
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Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl
Answer:
Mole fraction H₂O → 0.72
Mole fraction C₂H₅OH → 0.28
Explanation:
By the mass of the two elements in the solution, we determine the moles of each:
25 g . 1 mol/ 18g = 1.39 moles of water (solute)
25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)
Mole fraction solute = Moles of solute / Total moles
Mole fraction solvent = Moles of solvent / Total moles
Total moles = Moles of solute + Moles of solvent
1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles
Mole fraction H₂O = 1.39 / 1.933 → 0.72
Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28
Remember that sum of mole fractions = 1
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane