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sertanlavr [38]
2 years ago
5

50 POINTSSSSS PLZ HELPPPPPP NEED TO SHOW WORK!!!!

Chemistry
1 answer:
VashaNatasha [74]2 years ago
5 0
(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ 

<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>

<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>

<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>

<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>

<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>

<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>

<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
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Please Help Potassium sulfate has a solubility of 15g/100g water at 40 Celsius. A solution is prepared by adding 39.0g of potass
Svetradugi [14.3K]
To determine the state of saturation of the solution, we calculate the mass of solute per mass of water for the given amounts and compare this value to the solubility. If the value is less than the solubility, then the solution is unsaturated. If it is greater than solubility, then it is supersaturated. If it is equal to the solubility, then it is saturated.

mass solute / mass water  = 39.0 grams K2SO4 / 225 grams H2O = 0.173 g K2SO4/ g H2O
solubility = 15 g /100 g = .15 g/g

Therefore, the solution is supersaturated. When it is shaken, some of the solute would precipitate out. 

mass of solute soluble to water = .15 g K2SO4/ g water ( 225 g water ) = 33.75 g K2SO4
mass of K2SO4 that would crystallize = 39.0 - 33.75 = 5.25 g K2SO4
5 0
2 years ago
1. Complete the following Data Table as you conduct the lab
miskamm [114]

Answer:

Do to half of the mnairals this can not be made into a lab there is an error

Explanation:

4 0
2 years ago
Read 2 more answers
The element thallium is 70% thallium-205 and 30% thallium-203. Calculate its relative atomic mass to 1 decimal place.
xxMikexx [17]

Answer:

answer-

The relative atomic mass = 204.4

explanation:

Thallium -203 = 30%

Thallium -205 = 70%

Therefore ,

relative mass of thallium = (30×203 + 70×205)/100

relative mass of thallium = (20440)/100

relative mass of thallium = 204.40 amu

Thus,

relative atomic mass of thalium =204.4 ( to 1 decimal place)

6 0
2 years ago
Kaia, a chemical engineering graduate, has documented all titration procedures in her project report. She refers to this report
vagabundo [1.1K]

Answer:

The correct option is;

d. Explicit knowledge

Explanation:

Explicit knowledge is the knowledge that can be easily articulated documented stored in a retrieval system accesses, transmitted and shared with others

Tacit knowledge is the skill developed by an individual based on actual experience such that such knowledge comprise of both facts and perspectives

Hence explicit knowledge and tacit knowledge are complementary

The operations performed by Kaia include documentation, storing in a retrieval system (her project report) and accessing what she documented, this is an example of explicit knowledge.

7 0
2 years ago
What is the pH of a solution made by mixing 15.00 mL of 0.100 M HCl with 50.00 mL of 0.100 M KOH? Assume that the volumes of the
denis23 [38]

Answer:

The correct answer is: pH = 12.73

Explanation:

The <em>neutralization reaction</em> between HCl and KOH is given by the following chemical equation:

HCl + KOH ⇒ KCl + H₂O

Since HCl is a strong acid and KOH is a strong base, HCl is completely dissociated into H⁺ and Cl⁻ ions, whereas KOH is dissociated completely into K⁺ and OH⁻ ions.

For acids, the number of equivalents is given by the moles of H⁺ ions (in this case: 1 equivalent per mol of HCl). For bases, the number of equivalents is given by the moles of OH⁻ ions (in this case: 1 equivalent per mol of KOH).

The H⁺ ions from HCl will react with OH⁻ ions of KOH to give H₂O. The pH is calculated from the difference between the equivalents of H⁺ and OH⁻:

equivalents of H⁺= volume HCl x Molarity HCl

                            = (15.0 mL x 1 L/1000 mL) x 0.100 mol/L

                            = 1.5 x 10⁻³ eq H⁺

equivalents of OH⁻= volume KOH x Molarity KOH

                               = (50.0 mL x 1 L/1000 mL) X 0.100 mol/L

                               = 5 x 10⁻³ eq OH⁻

There are more OH⁻ ions than H⁺ ions. The excess of OH⁻ (that did not react with H⁺ ions) is calculated as follows:

OH⁻ ions= (5 x 10⁻³ eq OH⁻) -  (1.5 x 10⁻³ eq H⁺) = 3.5 x 10⁻³ eq OH⁻= 3.5 x 10⁻³ moles OH⁻  

As the volumes of the solutions are additive, the total volume of the solution is:

V= 15.0 mL + 50.0 mL = 65.0 mL= 0.065 L

So, the concentration of OH⁻ ions in the solution is given by:

[OH⁻] = moles OH⁻/V= (3.5 x 10⁻³ moles OH⁻)/0.065 L = 0.054 mol/L = 0.054 M  

From  [OH⁻], we can calculate pOH:

pOH = -log [OH⁻] = -log (0.054) = 1.27

Finally, we know that pH + pOH= 14; so we calculate pH:

pH= 14 - pOH = 14 - 1,27 =  12.73                                                            

8 0
3 years ago
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