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3 years ago

Which of the following four orbitals is FURTHEST from the nucleus?

Chemistry

1 answer:

coldgirl [10]3 years ago

7 0

Answer:

Explanation:

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What is the volume in L of 3.8 kg of this liquid?

DedPeter [7]

1L ------------ 1 kg
? ------------- 3.8 kg

3.8 x 1 / 1

= 3.8 L

4 0

2 years ago

In a chemical process, three bottles of a standard fluid are emptied into a larger container. A study of the individual bottles

Setler79 [48]

Answer:

The expected value is = 39 units

The standard deviation is = 0.1212 unites

The probability is = 0.2047 units

Explanation:

The explanation of this answer is shown on the first uploaded image

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3 years ago

Convert 125m^3 to km^3

Ugo [173]

Answer:

1.25e-7

Explanation:

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3 years ago

This is the equation for the combustion of propane.

yan [13]

Answer:

Explanation:

Reactants are the compounds which reaction produce the products. In general terms this can be expressed symbolically as follows:

reactants -> products

Other phenomena like heat are omitted because are not always present, that is, only compounds are included. Therefore, in this reaction the reactants are C3H8 (propane) and O2 ( oxygen) and the products are CO2 (carbon dioxide) and H2O (water)

8 0

2 years ago

Read 2 more answers

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Agata [3.3K]

<u>Answer:</u> The molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$PbI_2\rightleftharpoons Pb^{2+}+2I^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Pb^{2+}][I^-]^2$

We are given:

$K_{sp}=7.9\times 10^{-9}$

Putting values in above equation, we get:

$7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L$

Hence, the molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

3 0

3 years ago