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cricket20 [7]
3 years ago
9

Using the milligrams of ascorbic acid you entered above, the ratio of total sample volume to aliquot volume, and the total milli

grams of the Vitamin C tablet that you dissolved, calculate the mass of ascorbic acid in the Vitamin C tablet for each trial. Do this by scaling up to find the amount (mg) of ascorbic acid in your 250 mL flask Enter your calculated mass of ascorbic acid in the Vitamin C tablet, for each trial Be sure to enter your calculated mass in the corresponding order that you entered your milligrams of ascorbic acid The milligrams of ascorbic acid you entered for entry #1 previously should correspond to the mass of ascorbic acid that you enter for entry #1 here Ascorbic Acid (mg)
Entry # Volume (ml) Mass % of Ascorbic Acid
#1 9.650 3.84
#2: 7.500 2.99
#3: 9.500 3.78
Engineering
2 answers:
m_a_m_a [10]3 years ago
6 0
Using the milligrams of ascorbic acid you entered above, the ratio of total sample volume to aliquot volume, and the total milligrams of the Vitamin C tablet that you dissolved, calculate the mass of ascorbic acid in the Vitamin C tablet for each trial. Do this by scaling up to find the amount (mg) of ascorbic acid in your 250 mL flask Enter your calculated mass of ascorbic acid in the Vitamin C tablet, for each trial Be sure to enter your calculated mass in the corresponding order that you entered your milligrams of ascorbic acid The milligrams of ascorbic acid you entered for entry #1 previously should correspond to the mass of ascorbic acid that you enter for entry #1 here Ascorbic Acid (mg)
Entry # Volume (ml) Mass % of Ascorbic Acid
#1 9.650 3.84
#2: 7.500 2.99
#3: 9.500 3.782. 2 one is the answer
Dmitriy789 [7]3 years ago
3 0

Answer: 7.500 2.99

Explanation:

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2 years ago
If a motorist moves with a speed of 30 km/hr, and covers the distance from place A to place B
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105 km

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3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

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62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

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Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

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\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

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10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

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