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Vera_Pavlovna [14]
3 years ago
8

When measuring the volume of a liquid in a graduating cylinder, you measure_____.

Chemistry
2 answers:
BaLLatris [955]3 years ago
7 0
The average between the volume at the top of and the bottom of the meniscus
nlexa [21]3 years ago
6 0

Answer: from the bottom of the meniscus

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Osmium is the most dense element we know of. A 22 g sample of Osmium has a volume of 100 cL. Calculate the density of Osmium in
kakasveta [241]

Answer:

a. V = 1000 mL

b. Denisty = 0.022 g/mL

Explanation:

a.

First we need to convert the volume of the Osmium into mL. For that purpose we are given the conversion unit as:

1 mL = 0.1 cL

Hence, the given volume of Osmium will be:

V = Volume of Osmium = 100 cL = (100 cL)(1 mL/0.1 cL) = 1000 mL

<u>V = 1000 mL</u>

b.

The density of Osmium is given by the following formula:

Density = mass/Volume

Denisty = 22 g/1000 mL

<u>Denisty = 0.022 g/mL</u>

5 0
3 years ago
Which formula represents a nonpolar molecule containing polar covalent bonds h-h?
zhuklara [117]
<span>H2. Since the difference in electronegativity between two identical atoms is 0, the resulting molecule is non-polar.</span>
3 0
2 years ago
Anyone have a good tip, or app to help me with nomenclature???
ExtremeBDS [4]
Nomenclature of what?
3 0
3 years ago
HELP
Semenov [28]

C, 0.746 mol Ag.

1 mol Ag = 6.022 x 1023 atoms of Ag -> 4.49 x 1023 atoms of Ag x 1mol Ag/6.022 x 1023 atoms -> 0.746 mol Ag

7 0
3 years ago
What is the average yearly rate of change of carbon-14 during the first 5000 years?
erica [24]

Answer:

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year

Explanation:

Given that the mass of the carbon 14 at the start = 5 gram

At the end of 5,000 years we will have;

A = A_0 \times e^{-\lambda \times t}

Where

A = The amount of carbon 14 left

A₀ = The starting amount of carbon 14

e = Constant = 2.71828

T_{1/2} = The half life

\lambda = 0.693/T_{1/2}

t = The time elapsed = 5000 years

λ = 0.693/T_{1/2} = 0.693/5730 = 0.0001209424

Therefore;

A = 5 × e^(-0.0001209424×5000) = 2.7312 grams

Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left

The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams

The average yearly rate of change of carbon-14 during the first 5000 years  is therefore;

2.2688 grams/(5000 years) = 0.0004538 grams per year

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.

5 0
3 years ago
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