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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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The equivalence 
means that n-5 is a multiple of 12.
that is
n-5=12k, for some integer k
and so
n=12k+5
for k=-1, n=-12+5=-7
for k= 0, n=0+5=5 (the first positive integer n, is for k=0)
we solve 5000=12k+5 to find the last k
12k=5000-5=4995
k=4995/12=416.25
so check k = 415, 416, 417 to be sure we have the right k:
n=12k+5=12*415+5=4985
n=12k+5=12*416+5=4997
n=12k+5=12*417+5=5009
The last k which produces n<5000 is 416
For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,
thus there are 417 integers n satisfying the congruence.
Answer: 417
means that n-5 is a multiple of 12.
that is
n-5=12k, for some integer k
and so
n=12k+5
for k=-1, n=-12+5=-7
for k= 0, n=0+5=5 (the first positive integer n, is for k=0)
we solve 5000=12k+5 to find the last k
12k=5000-5=4995
k=4995/12=416.25
so check k = 415, 416, 417 to be sure we have the right k:
n=12k+5=12*415+5=4985
n=12k+5=12*416+5=4997
n=12k+5=12*417+5=5009
The last k which produces n<5000 is 416
For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,
thus there are 417 integers n satisfying the congruence.
Answer: 417