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How do you round the mass of the plactic cup that is 5 grams
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round it? It is already rounded to a single digit
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Ok, For Cup C
Mass of cup,water,stirrer: 55.15 Mass of sodium bicarbonate: 2.02g mass of citric acid: 0.77g total mass: 57.94g
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I think I must have answered this elsewhere.
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Ice cubes of total mass 6.00 grams have temperature -4.0°C. Then they are placed in 50.0 grams of water at 20.0°C in an
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To find the equilibrium temperature of the water inside the cup, we can use the principle of
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Is is possible to get 100% on a percent yield?
Now, For Cup A I had : Mass of cup water & stirrer: 67.51 mass of sodium
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With good technique and good equipment, it is possible to get very close to 100%. Your theoretical
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Plastic Cup B
Mass of cup, water and stirrer: 53.75g mass of sodium bicarbonate: 2.02g mass of citric acid: 1.52g total mass of
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. Show that the equivalence amount of citric acid for 2.00 g of sodium bicarbonate is 1.52 g. *** 1
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Ok, UMMMM I thought I was good.. Till now..
Ok, For Cup C Mass of cup,water,stirrer: 55.15 Mass of sodium bicarbonate: 2.02g mass
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You know I know how to calculate the TY and AY and %Y now, but I'm confused on the finding limiting
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For Cup C
Mass of cup,water,stirrer: 55.15 Mass of sodium bicarbonate: 2.02g mass of citric acid: 0.77g total mass: 57.94g mass
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I don't know why the delay in discussing/asking questions about limiting reagents but then I'm not
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PLZ HELP
A canvas shopping bag has a mass of 600 grams. When 5 cans of equal mass are put into the bag, the filled bag has a mass
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We can start by finding the mass of the cans alone. We know that the filled bag (with the cans
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A canvas shopping bag has a mass of 600 grams.
When 5 cans of equal mass are put into the bag, the filled bag has a mass of 4
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We can begin by setting up an equation using the given information. Let x be the mass of each can in
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An empty container has a mass of 600 grams. When 8 cans of equal mass are put into the container, the filled container has a
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We can start by subtracting the mass of the empty container (600 grams) from the mass of the filled
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The masses of 10 books are found to be in an arithmetic sequence. If their total mass is 13 kg and the lightest book has a mass
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Sn = n/2 (a+a+(n-1)d) If Tn is the nth term, S10 = 5(a+T10) 13000 = 5(400+T10) T10 = 2200 The
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