Answer:
2.52 g NaCl
Explanation:
(Step 1)
To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:
Molarity = moles / volume (L)
After you convert mL to L, you can plug the given values into the equation and simplify to find moles.
136.9 mL / 1,000 = 0.1369 L
Molarity = moles / volume
0.315 M = moles / 0.1369 L
0.0431 = moles
(Step 2)
Now, you can use the molar mass to convert moles to grams.
Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol
Molar Mass (NaCl): 58.443 g/mol
0.0431 moles NaCl 58.443 g
------------------------------ x ------------------- = 2.52 g NaCl
1 mole
Answer:
how to help people that helps
Answer:
The correct answer is "Secondary active transport".
Explanation:
Secondary active transport is a form of across the membrane transport that involves a transporter protein catalyzing the movement of an ion down its electrochemical gradient to allow the movement of another molecule or ion uphill to its concentration/electrochemical gradient. In this example, the transporter protein (antiporter), move 3 Na⁺ into the cell in exchange for one Ca⁺⁺ leaving the cell. The 3 Na⁺ are the ions moved down its electrochemical gradient and the one Ca⁺⁺ is the ion moved uphill its electrochemical gradient, because Na+ and Ca⁺⁺are more concentrated in the solution than inside the cell. Therefore, this scenario is an example of secondary active transport.
Answer:
See explanation
Explanation:
Extraction has to do with the separation of the components of a mixture by dissolving the mixture in a set up involving two phases. One phase is the aqueous phase (beneath) while the other is the organic phase (on top). The solvents used for the two phases must not be miscible. Water commonly is used for the aqueous phase.
Ethanol is an important solvent in chemistry but the solvent is miscible with water in all proportions. As a result of this, ethanol is a poor solvent for carrying out extraction.
Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
