Answer: An object has a weight of 10 kg on the surface of Earth. If the same object were transported to the surface of Mars, the object would have a weight of 3.8 kg. The density of the object is greater on Earth than it is on Mars.
Explanation:
Answer:
0.66 × 10⁴² mol
Explanation:
Given data:
Moles of water = ?
Number of molecules of water = 3.99× 10⁶⁵ 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,
1 mole = 6.022 × 10²³ molecules of water
3.99× 10⁶⁵ molecules × 1 mol / 6.022 × 10²³ molecules
0.66 × 10⁴² mol
Answer:
1.02 moles of AlCl3
Explanation:
Consider the balanced reaction equation shown below;
2Al(s) + 3CuCl2(aq) -------> 2AlCl3 (aq) + 3Cu(s)
This is the balanced reaction equation for the reaction going on above. Recall that the first step in solving any problem is to accurately put down the chemical reaction equation. This balanced reaction equation always serves a reliable guide in solving the problem at hand.
Given that;
3 moles of CuCl2 yields 2 moles of AlCl3
1.53 moles of CuCl2 yields 1.53 × 2 / 3 =1.02 moles of AlCl3
Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
The answer is complex carbohydrates.
Hope that helps! :)
