The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4
1.00*10^3
You’d need to lower the exponent because rounding to 3 sig figs changes the 9’s to - 1000. Keep the 0’s.
A solution is turning blue means, it is turning it's behavior to Basic from Acidic. Reagent D must be a Basic component so it is increasing the pH of the solution. As reaction does not depend on the phase of the component, it could be anything i.e., Solid, Liquid or gas.
Hope this helps!
Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
Answer:
34.9 g of Zn(OH)₂ is the maximum mass that can be formed
Explanation:
Let's state the reaction:
ZnO(s) + H₂O(l) → Zn(OH)₂ (aq)
First of all, we need to determine the moles of each reactant and state the limiting:
28.6 g . 1mol /81.38 g = 0.351 moles of ZnO
9.54 g . 1mol /18 g = 0.53 moles of water
As ratio is 1:1, for 0.53 moles of water, we need 0.53 moles of ZnO, but we only have 0.351, so the limiting reactant is the ZnO.
Ratio with the product is also 1:1. From 0.351 moles of oxide we can produce 0.351 moles of hydroxide. Let's calculate the mass:
0.351 mol . 99.4 g /1mol = 34.9 g