The law of conservation of energy and the law of conservation of mass apply to chemical reactions

Why do igneous rocks not contain fossils in them ?

Mashutka [201]

Answer:

Igneous rocks do not contain any fossils.

Explanation:

This is because any fossils in the original rock will have melted when the rock melted to form magma. Sedimentary rocks are formed from the broken remains of other rocks that become joined together.

6 0

3 years ago

Based on the parameters set forth in your lab procedure, what is the ph you should observe when you added the hcl to water?.

Rzqust [24]

Acid has a pH below 7 while water has a pH of 7. A strong acid with a pH of roughly 3, HCl is. Water gets more acidic and loses pH in the range of 4-5 when HCl is added to it.

A substance's pH is a gauge of how basic or acidic it is. It is a measurement of the amount of H+ present in the solution. It is equivalent to the negative logarithm of the concentration of H+ ions mathematically.

A solution is acidic if its pH value is less than 7, and basic if it is greater than 7. Acids have a lower pH because they contain more H+ ions. Strong acid hydrochloric acid has a pH between 2 and 3.

Water has no charge. Water becomes acidic in pH if any acid is introduced. Therefore, if HCl is given to water, the pH of the water will change to 3-5 depending on the acid content.

To find more on pH refer here:

brainly.com/question/491373

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6 0

1 year ago

How would you make a 2.00L of 0. 500M sodium chloride solution. (assume you have a fully equipped lab with water); sketch, calc

Gwar [14]

Answer:

Sodium chloride solution:

First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.

Sulfuric acid dilution:

First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.

Explanation:

Sodium chloride solution:

Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.

Sulfuric acid dilution:

This is the equation for dilution of solutions:

$c_{1} v_{1} =c_{2} v_{2}$

Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.

When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:

$v_{1} =\frac{c_{2} v_{2} }{c_{1} }$

in this case that would be:

$v_{1} =\frac{0.100 x2.0 }{12.0 }=0.0166L=16.6mL$

5 0

3 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

4 years ago

Arthur is performing a lab which requires a 0.2 M solution of calcium hydroxide. However, the only calcium hydroxide solution av

aev [14]

Answer:

He needs to add 4 mL of the 0.5 M solution to 6 mL of water.

7 0

4 years ago