Answer:
8.7 mol Al
Explanation:
You have the compound Al₂Si₄O₁₀(OH)₂. Within the compound, you have atoms. You can tell how many atoms there are by looking at the subscript. You have 2 aluminum atoms, giving you 8.7 moles of aluminum.
4.35 × 2 = 8.7
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡ HI ≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴
Answer:
Element
Explanation:
An element is the simplest form of a matter. It can not be broken down into a simpler substance without loss of identity.
∞║║ HOPE THIS HELPS PLEASE MARK AS BRAINLIEST║║∞
Hey there !
Given the reaction:
N2 + 3 H2 = 2 NH3
At constant pressure and temperature ,volume is proporcional to moles:
Theoretical moles of N2 and H2 => 1:3
Theoretical volume of N2 and H2 => 1:3
Experimental volume of N2 and H2 => 3.0 L : 4.0 L
0.75 : 1 = 2.25 : 3
Since N2 is in excess reactant , H2 is the limiting reactant
Therefore:
volume of NH3 is 2/3 * Volume of H2
= 2/3 * 4.0 = 2.66 L
Hope that helps!
Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:
First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?
= 25000 g
Substitute the values into the formula:
= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.
This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:
H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O
So, we do the stoichiometric calculations. The important data we should know is the molar mass of oxalic acid which is equal to 90 g/mol.
(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%
This is impossible. The purity can't be more than 100%. Looking at our calculations and the balance reaction, all steps were done correctly. So, I think there is some typographical error in the given. The mass of the sample should be 7.984 g. Then, the answer would be 48.87% purity.
