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Scrat [10]
3 years ago
7

G vinegar is a solution of acetic acid in water. if a 165 ml bottle of distilled vinegar contains 28.2 ml of acetic acid, what i

s the volume percent (v/v) of the solution?
Chemistry
1 answer:
Delicious77 [7]3 years ago
7 0
<span>Volume Percent (volume/volume%) is defined to be the ratio of the volume of solute to the volume of solution times 100%. The acetic acid is the solute volume. The bottle of vinegar is the solution volume. Acetic acid (28.2ml) / Vinegar(165ml) x 100% = 17.1%</span>
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2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??
Zepler [3.9K]

Answer:

d=1.78\times 10^{-7}\ Dg/\mu m^3

Explanation:

Given,

The density of Helium is 1.78\times 10^4\ g/cm^3

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3

So, the density of Helium is equal to 1.78\times 10^{-7}\ Dg/\mu m^3.

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2 years ago
How might a cave an ant and a lake each meet the needs of an oranism
PIT_PIT [208]
The organism could live in the cave, eat the ant, and drink from the lake.
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3 years ago
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satela [25.4K]

Answer:

There is a lot of empty space between them

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We know that gas molecules are loosely packed,

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3 0
3 years ago
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Determine the number of grams of C4H10 that are required to completely react to produce 8.70 mol of CO2 according to the followi
Nitella [24]

Answer:

126.4 g of C_{4}H_{10} are required

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8 moles of CO_{2} are produced from 2 moles of C_{4}H_{10}

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Molar mass of C_{4}H_{10} = 58.12 g/mol

So, mass of C_{4}H_{10} required = (2.175\times 58.12)g = 126.4 g

Hence 126.4 g of C_{4}H_{10} are required

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