The answer is phosphorus :))
Answer:
5.35×10²⁴ atoms of Li
Explanation:
To solve this we apply this rule of three:
We know that 1 mol contains 6.02×10²³ atoms
Therefore, 8.9 moles must contain (8.9 . 6.02×10²³) / 1 = 5.35×10²⁴ atoms
One mol is the amount of substance containing as many elemental entities as atoms, which are contained in 12 g of the C¹² isotope
In conclussion 6.02×10²³ atoms of C¹² isotope weighs 12 g
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.
The new volume of the gas is 144 mL.
<u>Explanation:</u>
At STP, the temperature is 273 K an pressure is 1 atm.
1 atm=101kPa
Given that initial volume V1=36 mL
Initial pressure P1=1atm=101 kPa
Final pressure P2=25.3kPa
We have to determine the final volume V2
According to Boyle's law
=144 mL
Final volume =144mL