Solutions are a type of mixture.
A solution is a homogeneous mixture of particles so small that they cannot be seen and cannot be filtered out.
Definition: Homogeneous mixture looks the same.
I grabbed these from my science notes, hope this helps.
Answer:
glucose is in large sugar molecules, while fructose is in simple fruits and vegetables
Answer:
0.1056 mole
Explanation:
As Sally knows that the charge on the metal ion is n = +2
In that compartment ![$[M^{n+}]=[m^{2+}]=8.279 \ M$](https://tex.z-dn.net/?f=%24%5BM%5E%7Bn%2B%7D%5D%3D%5Bm%5E%7B2%2B%7D%5D%3D8.279%20%5C%20M%24)
The volume of the 
taken in that compartment = 6.380 mL
So, the number of moles of 
= 52.82 m mol
= 0.05280 mol
But n = 2
Therefore, moles of 
= 2 x moles of 
= 2 x 0.05282
= 0.1056 mole
Answer:
3). 1.30 × 10^(24) molecules
Explanation:
From avogadro's law which state that equal volume of all gases at the same temperature and pressure contain the same number of molecules.
We can relate it to this question as;
V₁/n₁ = V₂/n₂
Where;
V₁ is initial volume
n₁ is initial number of molecules
V₂ is final volume
n₂ is final number of molecules
Thus at STP, we have V₁ = V₂ and as such Plugging in the relevant values gives;
5/(1.30 x 10^(24)) = 5/n₂
n₂ = 1.30 x 10^(24) molecules
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
