Answer:
Scientists were unsure about the last common ancestor between apes and humans because fossils are so scarce, researchers do not know what the last common ancestors of living apes and humans looked like or where they originated.
Explanation:
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
The Photo Isn't Clear At All.
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
Answer:
Will be more acidic
Explanation:
The equilibrium of NH3 in water is:
NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq).
Where equilibrium constant, Kb, is:
Kb = 1.85x10⁻⁵ = [NH4⁺] [OH⁻] / [NH3]
From 0.10M NH3, the reaction will produce X of NH4⁺ and X of OH⁻ and Kb will be:
1.85x10⁻⁵ = [X] [X] / [0.10M]
1.8x10⁻⁶ = X²
X = 1.34x10⁻³ = [OH⁻]
As pOH = -log[OH⁻] = 2.87
And as pH = 14 - pOH
pH of the 0.10M NH3 is 11.13
Now, to find the pH of the NH4Cl and NH3 we need to use H-H equation for bases:
pOH = pKb + log [NH4⁺] / [NH3]
<em>Where pKb is -log Kb = 4.74 and [] are moles of both compounds.</em>
<em />
Moles of [NH4⁺] = [NH3] = 60mL, 0.060L*0.22M = 0.0132moles:
pOH = 4.74 + log [0.0132] / [0.0132]
pOH = 4.74
pH = 14 - 4.74 = 9.26
That means the pH of the resulting solution will be more acidic
