1. You can use Avogadro’s number, 6.022x10^23 atoms/mole, to answer this one.
(3.311x10^24/6.022x10^23) = 5.498 moles of substance
2. H2O has a formula weight approximately equal to 18 grams. Dividing the given amount by the formula weight of water will tell us the number of moles present.
126/18 = 7 moles H2O
This question belongs to physics subject.
The gravitational attraction is ruled by the Universal Law of Gravity (Newton's Law).
This law states the the atracction force between two objects is proportional to the product of the masse of the two objects.
The if one of the masses is doubled the product of the masses will also double, and the resultant force will be the double.
Then, the answer 2*22N = 44N is the right option (option C)
H2O(s) ==> H2O(l)
<span>The density of liquid H2O = 1.00 g/mL. </span>
<span>The density of ice is 0.917 g/mL. </span>
<span>Suppose we start with 1 mol (18 g) H2O(s) and it melts to the liquid. </span>
<span>As a liquid it will occupy 18g/1.00 = 18 mL. </span>
<span>As a solid it will occupy 18/0.917 = 19.63 mL. </span>
<span>19.63-18.0 = 1.63 mL diffeence. </span>
<span>The heat fusion of ice is 6.01 kJ/mol; therefore, 6.01/1.63 = about 3.69 kJ for each mL difference. </span>
<span>In this experiment we have a difference of 0.109 mL; then 3.69 kJ/mL x 0.109 mL = 0.402 kJ for 0.0657 g Zn. You want per mol Zn and you have 0.0657/65.39 = about 0.001 mol Zn so </span>
0.402 kJ/0.001 = <span>402 kJ/mol Zn</span>
Answer: Your balanced equation is 2Al + 3S --> Al_2S_3. In two separate equations, first use the 48.5g of Al to find the amount (in moles) of Al_2S_3. Then use the 62.8g of S to find the amount (in moles) of Al_2S_3. By doing these, you find which reactant (the Al or the S) produces the least amount of Al_2S_3. Since it's the S, that's the limiting reactant: you then use that 62.8g of S to find how many grams of Al will be used in the reaction. (Use the same balanced equation.) You find that 35.2g of Al will be used, then find the excess by subtracting that 35.19 from the original 48.5g. Your excess amount is 13.3g of Al!
Explanation:
hopefully this is what you were looking for <3 :)
The reaction is
Zn(s) + 2MnO₂(s) + H₂O(l) ⟶ Zn(OH)₂(s) + Mn₂O₃(s)
The reactants are Zn(s) and MnO₂(s) while Zn(OH)₂(s) and Mn₂O₃(s) products.
Let's see the oxidation state of each compound.
sum of the o.s. of the each element of compound = charge of the compound
O.s of O = -2
O.s of OH⁻ = -1
O.s of Zn(s) = 0
O.s of Mn in MnO₂(s) = x
x + (-2) * 2 = 0
x = +4
O.s of Zn in Zn(OH)₂(s) = a
a + (-1) * 2 = 0
a = +2
O.s of Mn in Mn₂O₃(s) = b
2*b + (-2) * 3 = b
2b = 6
b = +3
Since the O.s is reduced in Mn from +4 to +3, the reduced species is Mn₂O₃(s).