Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
The correct answer choice from the statement above about investigated factors from lab results is :
Students who spend less time studying after school get lower math grade
Option D is the correct answer
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So therefore, The correct answer choice from the statement above about investigated factors from lab results is :
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Answer:
Option b, Irreversible hydrocolloid impression material
Explanation:
Irreversible is used as an impression material to take impression from edentulous jaws. It is also used in wound healing and drug delivery.
Alginate is a natural polymer found in cell wall of brown seaweed.
Its monomers are β-D-mannuronate and α-L-guluronate.
In association with Ca2+, it forms gel. It is hydrophilic in nature.
<u>Answer:</u>
The correct answer option is a) collisions between the particles and surrounding molecules.
<u>Explanation:</u>
The collisions between the particles and surrounding molecules causes the Brownian motion of particles in a colloid.
Brownian motion is the irregular movement of the microscopic particles in a fluid which bombard into each other.
It basically is the result of the molecules of a dispersion medium colliding with the dispersed particles of the phase.
Answer:
The correct answer is 8.786 g CaCO₃
Explanation:
The balanced reaction is the following:
CaCl₂(ac) + K₂CO₃(ac) → CaCO₃(s) + 2 KCl(ac)
From the stoichiometry, 1 mol of CaCl₂ (111 g) reacts with 1 mol of K₂CO₃ (138 g) to form 1 mol CaCO₃(100 g) and 2 moles of KCl (149 g).
The stoichiometric ratio CaCl₂/K₂CO₃ is: 111 g/138 g= 0.80 g CaCl₂/K₂CO₃.
We have 14.584 g CaCl₂ and 12.125 g K₂CO₃, which gives a ratio of: 14.584g/12.125 g= 1.2 g CaCl₂/K₂CO₃.
0.8 ∠ 1.2 ⇒ K₂CO₃ is the limiting reactant
We use the limiting reactant to calculate the grams of CaCO₃ produced. For this, we know that from 138 g K₂CO₃ 100 g of CaCO₃ are produced. So, we multiply the amount of K₂CO₃ by this stoichiometric ratio to obtain the grams of CaCO₃ produced:
12.125 g K₂CO₃ x 100 g CaCO₃/138 g K₂CO₃= 8.786 g CaCO₃
Therefore, the theoretical yield of CaCO₃ is 8.786 g.