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3 years ago

Which statement best describes a physical property?

Chemistry

1 answer:

ArbitrLikvidat [17]3 years ago

3 0

The first I believe

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The magnitude of an earthquake can range from less than 1 to at least

shusha [124]

The magnitude of an earthquake can range from 1 to at least 8

6 0

3 years ago

Write the balanced net ionic equation for the reaction between mno4- and fe2+

Citrus2011 [14]

2MnO4 - + Fe2+ ------> Fe2(MnO4)2

7 0

4 years ago

On a mission to a newly discovered planet, an astronaut finds chlorine abundances of 13.85 % for 35Cl and 86.15 % for 37Cl. What

topjm [15]

Answer : The atomic mass of chlorine is, 36.723 amu

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:

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

As we are given that,

Mass of 35-Cl isotope = 35 amu

Percentage abundance of 35-Cl isotope = 13.85 %

Fractional abundance of 35-Cl isotope = 0.1385

Mass of 37-Cl isotope = 37 amu

Percentage abundance of 37-Cl isotope = 86.15 %

Fractional abundance of 37-Cl isotope = 0.8615

Now put all the given values in above formula, we get:

$\text{Average atomic mass }=[(35\times 0.1385)+(37\times 0.8615)]$

$\text{Average atomic mass }=36.723amu$

Therefore, the atomic mass of chlorine is, 36.723 amu

6 0

4 years ago

I really kinda need some hell right now :)

Fudgin [204]

It’s c 1.59 x 10^4L i’m positive

4 0

3 years ago

You need to make an aqueous solution of 0.237 M chromium(III) acetate for an experiment in lab, using a 500 mL volumetric flask.

julsineya [31]

Answer: 27.1 g of solid chromium(III) acetate should be added.

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Cr(CH_3COO)_3$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{229g/mol}$

Now put all the given values in the formula of molality, we get

$0.237=\frac{xg\times 1000}{229g/mol\times 500ml}$

$x=27.1$

Therefore, 27.1 g of solid chromium(III) acetate should be added.

3 0

3 years ago