An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of 55 degrees C is mixed with the water in the insulated container. When thermal equilibrium is reached, the temperature of the mixture is 25 degrees C. Assume that heat flows only between the alloy and the water. What is the specific heat of the alloy?
Answers
Answer:
The specific heat of the alloy is 2.324 J/g°C
Explanation:
Step 1: Data given
Mass of water = 0.3 kg = 300 grams
Temperature of water = 20°C
Mass of alloy = 0.090 kg
Initial temperature of alloy = 55 °C
The final temperature = 25°C
The specific heat of water = 4.184 J/g°C
Step 2: Calculate the specific heat of alloy
Qlost = -Qwater
Qmetal = -Qwater
Q = m*c*ΔT
m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)
⇒ mass of alloy = 90 grams
⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED
⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C
⇒ mass of water = 300 grams
⇒ c(water) = the specific heat of water = 4.184 J/g°C
⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C
90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C
c(alloy) = 2.324 J/g°C
The specific heat of the alloy is 2.324 J/g°C
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Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated steam. (a) Given these data, find ΔH o rxn , ΔG o rxn , and ΔS o rxn at 298 K: ΔH o f (kJ/mol) ΔG o f (kJ/mol) S o (J/mol·K) Ethylbenzene, C6H5−CH2CH3 −12.5 119.7 255 Styrene, C6H5−CH=CH2 103.8 202.5 238 ΔH o rxn ΔG o rxn ΔS o rxn kJ kJ J/K (b) At what temperature is the reaction spontaneous? °C (c) What are ΔG o rxn and K at 600.°C? ΔG o rxn K kJ/mol × 10 Enter your answer in scientific notation.
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Does dissolved potassium chloride affect the surface tension between water molecules?
Answers
Answer:
The correct answer is 5.6 × 10⁻²³ M.
Explanation:
As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.
Let the solubility of Hg₂Br₂ is S mol per liter.
Therefore,
KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)
Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)
Net [Br-] = (2S + 1) M
Ksp = S (2S + 1)²
Ksp = S (4S² + 1 + 4S)
Ksp = 4S³ + S + 4S²
As the solubility is extremely less, therefore, we can ignore S² and S³. Now,
Ksp = S = 5.6 × 10⁻²³ M
Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.
Answers
Answer:
Saturated.
Explanation:
Hello,
Animal fats are lipids derived from animals which are commonly solid at room temperature and mainly constituted by triglycerides which are strictly chemically saturated with hydrogen, it means they do not tend to have double or triple bonded carbon atoms but just single-bonded carbons. This fact suggests that animal fats provide more energy than vegetable fats because they have more C-H bonds that when broken increase the total provided energy.
Best regards.
Answers
Answer:
Explanation:
Hello,
In this case, for strontium phosphate, we find an ionic equation for its dissociation as shown below:
Next, the solubility product is found by applying the law of mass action, considering that the solid salt is not considered but just the aqueous species due to heterogeneous equilibrium:
Best regards.
Answers
Answer:
Tinitial = 78.6°C
Explanation:
In a calorimetry experiment, the flow heat is measured for a system that has a state change. In this case, there isn't happening a physical change, so the heat is called sensitive heat, and it's calculated by:
q = mxCpxΔT
Where q is the heat, m is the mass, Cp is the specific heat and ΔT is the difference of final and initial temperature (Tfinal - Tinitial).
Copper is losing heat, so q is negative, then:
-1860 = 92x0.377x(25 - Ti)
34.684(25 - Ti) = -1860
25 - Ti = -53.63
-Ti = -78.63
Ti = 78.6ºC
Answers
Explanation:
Ionic equation
NaCl(aq) --> Na+(aq) + Cl-(aq)
Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)
In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.
Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)
= 142 g/mol
Molecular weight of NaCl = 23 + 35.5
= 58.5 g/mol
Masses
% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100
= 46/142 * 100
= 32.4%
% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100
= 23/58.5 * 100
= 39.3%
Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.
Final answer:
You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.
Explanation:
No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.
For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.
Learn more about Chemical Substitution here:
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completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL
Answers
Answer:
B) 125 mL
Explanation:
M1V1=M2V2
(0.120M)(x)=(150.0 mL)(0.100M)
x= 125 mL
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