Answer:
1.02 grams
Explanation:
The reaction between Na₃PO₄ and BaCl₂ is:
2Na₃PO₄(aq) + 3BaCl₂(aq) → Ba₃(PO₄)₂(s) + 6NaCl(aq)
We need to find the number of moles of Na₃PO₄ and BaCl₂:
![n_{Na_{3}PO_{4}} = C*V = 0.200 M*0.050 L = 0.01 moles](https://tex.z-dn.net/?f=%20n_%7BNa_%7B3%7DPO_%7B4%7D%7D%20%3D%20C%2AV%20%3D%200.200%20M%2A0.050%20L%20%3D%200.01%20moles%20)
![n_{BaCl_{2}} = C*V = 0.100 M*0.050 L = 0.005 moles](https://tex.z-dn.net/?f=%20n_%7BBaCl_%7B2%7D%7D%20%3D%20C%2AV%20%3D%200.100%20M%2A0.050%20L%20%3D%200.005%20moles%20)
Now, we need to find the limiting reactant:
![n_{Na_{3}PO_{4}} = \frac{2}{3}*0.005 moles = 0.0033 moles](https://tex.z-dn.net/?f=%20n_%7BNa_%7B3%7DPO_%7B4%7D%7D%20%3D%20%5Cfrac%7B2%7D%7B3%7D%2A0.005%20moles%20%3D%200.0033%20moles%20)
We have that it is needed 0.0033 moles of Na₃PO₄ to react with BaCl₂ and we have 0.01 moles of Na₃PO₄, so the limiting reactant is BaCl₂.
The number of moles of Ba₃(PO₄)₂ is:
![n_{Ba_{3}(PO_{4})_{2}} = \frac{0.005}{3} = 0.0017 moles](https://tex.z-dn.net/?f=%20n_%7BBa_%7B3%7D%28PO_%7B4%7D%29_%7B2%7D%7D%20%3D%20%5Cfrac%7B0.005%7D%7B3%7D%20%3D%200.0017%20moles%20)
Finally, the mass of Ba₃(PO₄)₂ is:
![m = n_{Ba_{3}(PO_{4})_{2}}*M = 0.0017 moles*601.93 g/mol = 1.02 g](https://tex.z-dn.net/?f=%20m%20%3D%20n_%7BBa_%7B3%7D%28PO_%7B4%7D%29_%7B2%7D%7D%2AM%20%3D%200.0017%20moles%2A601.93%20g%2Fmol%20%3D%201.02%20g%20)
Therefore, should be obtained 1.02 grams of the precipitate.
I hope it helps you!
Answer:
Star B
Explanation:
Star B, this is because, usually in the discussions about stars, the more luminous star or say, the brighter star usually is the one that possesses the smaller absolute brightness.
That being said, succinctly put, if we have two stars A and star B, if star A is brighter than star B, then star A has a lower absolute brightness. Also, if star B is less brighter than star A, then it has the higher absolute brightness.
False, different density of surroundings gives different speeds
Answer:
The correct answer is option B and D, that is, halogen (chlorine) and hydroxyl.
Explanation:
An artificial sweetener and sugar substitute is sucralose. It is noncaloric as the majority of the sucralose ingested does not get dissociated within the body. The generation of sucralose takes place by the chlorination of sucrose. It is about 300 to 1000 times sweeter in comparison to sucrose.
The consumption of sucralose is safe for both nondiabetics and diabetics, it is used in various food and beverage components due to non-caloric sweetener characteristics. It does not affect the levels of insulin and does not affect dental health. As it is produced by chlorination of sucrose, thus, the functional groups present in it are a halogen (chlorine) and a hydroxyl.