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Matt did an experiment to study the solubility of two substances. He poured 100 mL of water at 20 °C into each of two beakers labeled A and B. He put 50 g of Substance A in the beaker labeled A and 50 g of Substance B in the beaker labeled B. The solution in both beakers was stirred for 1 minute. The amount of substance left undissolved in the beakers was weighed. The experiment was repeated for different temperatures of water and the observations were recorded as shown.
Part 1: Which, if any, substance is soluble in water?
Part 2: Explain how the data helped you determine solubility for both substances for temperatures 20 °C to 80 °C.
Matt did an experiment to study the solubility of two substances. He poured 100 mL of water at 20 °C into each of two beakers labeled A and B. He put 50 g of Substance A in the beaker labeled A and 50 g of Substance B in the beaker labeled B. The solution in both beakers was stirred for 1 minute. The amount of substance left undissolved in the beakers was weighed. The experiment was repeated for different temperatures of water and the observations were recorded as shown.
Part 1: Which, if any, substance is soluble in water?
Part 2: Explain how the data helped you determine solubility for both substances for temperatures 20 °C to 80 °C.
1 answer:
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a. ×
A
b. ×
A
c. A
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
<h3>How to determine the current</h3>The formula for finding current
I = V/R
Where v = voltage
R = resistance
A. V = 12V
R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series
×
A
B. V = 9V
R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series
×
A
C. V= 12V
1/R = =
×
R = = 310. 56 Ω
A
It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
Learn more about Ohms law here:
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