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3 years ago

In scientific notation, 213,000 is equal to

Chemistry

2 answers:

kirza4 [7]3 years ago

5 0

2.13 x 10^5 I think.

g100num [7]3 years ago

4 0

<span>2.13 x 10</span>⁵ is 213,000 in scientific notation.

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A sample of gold has a mass of 645.0 lb and a volume of 0.5348 ft. What is its specific gravity? Number SG= What are the units o

LekaFEV [45]

Explanation:

The given data is as follows.

Mass of gold = 645 lb, Volume = 0.5348 ft3

Density of water = 62.4 lbs/ft3

It is known that specific gravity is defined as density of substance divided by the density of standard fluid.

Mathematically, Specific gravity = $\frac{\text{density of gold}}{\text{density of water}}$

Specific gravity = $\frac{\text{density of gold}}{62.4 (lbs.ft^{-3})}$

Now, calculate the density of gold then from density we will calculate specific gravity as follows

Since, Density = $\frac{mass}{volume}$

Density = $\frac{645 lbs}{0.5348 ft^{3}}$

= 1206.06 $lbs/ft^{3}$

As, Specific gravity = $\frac{\text{density of gold}}{\text{density of water (standard fluid)}}$

= $\frac{1206.06 (lbs/ft^{3})}{62.4 (lbs/ft^{3})}$

= 19.32

Therefore, the value of specific gravity is 19.32.

Specific gravity has no units as it is density divided by density. Hence, all the units get canceled out.

7 0

3 years ago

The average molecular weight for element X is 59.97 g/mol. There are two known isotopes of element X, one weighing 59 g/mol, and

Temka [501]

Answer:

The correct answer is:48.5% X-59, 51.5% X-61

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of X-59 isotope be 'x'. So, fractional abundance of X-61 isotope will be '1 - x'

For X-59 isotope:

Mass of X-59 isotope =59 g/mol

Fractional abundance of X-59 isotope = x

For X-61 isotope:

Mass of X-61 isotope = 61 g/mol

Fractional abundance of X-61 isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$59.97 g/mol=[(59 g/mol\times x)+(61 g/mol\times (1-x))]\\\\x=0.515$

Percentage abundance of X-59 isotope = $0.515\times 100=51.5\%$

Percentage abundance of X-61 isotope = $(1-0.515)=0.485\times 100=48.5\%$

Hence, the percentage abundance of both the isotopes X-59and X-61 are 51.5% and 48.5% respectively.

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3 years ago

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notsponge [240]

Answer:

1. B. environment

2. A. true

Explanation:

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andrezito [222]

95.611 g/mol that's the answers

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densk [106]

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4 years ago

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