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3 years ago

What is bed rock made of

Chemistry

2 answers:

Darya [45]3 years ago

8 0

Answer:

Bedrock is the hard, solid rock beneath surface materials such as soil and gravel.

Bedrock can be made of most types of rock, such as granite, limestone, or sandstone.

Hope this helps!

brilliants [131]3 years ago

4 0

Bedrock can be made of most types of rock, such as granite, limestone, or like this piece of bedrock, sandstone. Bedrock is the hard, solid rock beneath surface materials such as soil and gravel.

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Can you store cuso4 in an aluminum container? explain your reasoning

mash [69]

We can store the copper sulphate solution in alumiun container, if cover on alumiun is present.

<h3>Can you store cuso4 in an aluminum container?</h3>

Aluminium is more reactive than copper so the Aluminium will displace copper sulphate from its solution by reacting with it but if there is cover on the aluminium then the alumium can't react with copper.

So we can store the copper sulphate solution in alumiun container.

Learn more about container here: brainly.com/question/11459708

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2 years ago

Which type of intermolecular attractive force operates between (part A) all molecules, (part B) polar molecules, (part C) the hy

nydimaria [60]

Answer:

PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.

PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.

PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.

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3 years ago

How can you differentiate the Atomic Structure and the Bohr Model? In other words, how is the Atomic Structure different from th

Tatiana [17]

Answer:

Bohr thought that electrons orbited the nucleus in circular paths; whereas in the modern view atomic electron structure is more like 3D standing waves. Bohr built upon Rutherford's model of the atom. ... He believed that electrons moved around the nucleus in circular orbits with quantised potential and kinetic energies.

Explanation:

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3 years ago

A lab is done using different metals and the density of the metals are recorded in a data table. What is the independent variabl

vovikov84 [41]

Answer: The independent variable is the type of metal being used.

{Note: The "dependent variable" is the "measured density" that corresponds to each of the metals."}.

___________________________________________

Explanation:

___________________________________________

The "independent variable", which is plotted on the "x-axis" (horizontal axis), is the variable that can be "controlled/manipulated". In this case, this would be the type of metal chosen.

The "dependent variable" , which is plotted on the "y-axis" (vertical axis) is the "obtained value/measurement/result" (that "cannot be controlled/manipulated").

In this case, the "density", which is the "measured value" that corresponds to the selected "meal", is the "dependent variable".

___________________________________________

Hope this helpful to you!

Wishing you well!

___________________________________________

6 0

3 years ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

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

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

4 0

3 years ago