B) differences in temperature
Explanation:
The factor that most directly affects the flow of ocean currents is the differences in ocean temperature from one place to another.
Ocean current is the directional movement of the ocean water in its basin.
The main cause of ocean current is differences in temperature.
- Ocean water on the surface receives direct sunlight.
- This warms and the surface and decreases their density.
- When winds blows over them, they cool and sink under their own weight.
- The less cold water in depths rise to replace them.
- This exchange of ocean water due to loss and gain of heat cause ocean current.
learn more:
Ocean density brainly.com/question/6760255
#learnwithBrainly
<h3>Answer:</h3>
#1. Ca²⁺
# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
<h3>Explanation:</h3>
The question above concerns solubility of salts or ions in water.
The solution given contains Ag+, Ca2+, and Co2+ ions.
- In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
- In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
- The net ionic equation for the reaction is;
Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
- From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
- In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
- The net ionic equation for the reaction is;
3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
- According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
The first element discovered through synthesis was technetium<span>—its discovery being definitely confirmed in 1936. Hope that helps.</span>
Answer:
20ppm
Explanation:
parts per million are defined as the mass of solute in mg (In this case, mass of DDT) per kg of sample.
To solve this question we must find the mass of DDT in mg and the mass of sample in kg:
<em>Mass DDT:</em>
0.10g * (1000mg / 1g) = 100mg
<em>Mass sample:</em>
5000g * (1kg / 1000g) = 5kg
Parts per Million:
100mg / 5kg =
<h3>20ppm</h3>
