Answer:
La temperatura final del sistema es 1029,346 °C.
Explanation:
Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:
[tex]m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)[/tex]
[tex](m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}[/tex]
[tex]T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}}[/tex] (1)
Donde:
[tex]m_{Sn}[/tex] - Masa del estaño, en gramos.
[tex]m_{Cu}[/tex] - Masa del cobre, en gramos.
[tex]c_{Sn}[/tex] - Calor específico del estaño, en calorías por gramo-grados Celsius.
[tex]c_{Cu}[/tex] - Calor específico del cobre, en calorías por gramo-grados Celsius.
[tex]T_{Sn}[/tex] - Temperatura inicial del estaño, en grados Celsius.
[tex]T_{Cu}[/tex] - Temperatura inicial del cobre, en grados Celsius.
Si sabemos que [tex]m_{Cu} = 150\,g[/tex], [tex]m_{Sn} = 35\,g[/tex], [tex]c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]T_{Sn} = 560\,^{\circ}C[/tex] y [tex]T_{Cu} = 1100\,^{\circ}C[/tex], entonces la temperatura final del sistema es:
[tex]T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}[/tex]
[tex]T = 1029,346\,^{\circ}C[/tex]
La temperatura final del sistema es 1029,346 °C.
1. What is the equivalent pressure measurement in mmHg of 2.50 atm?
Answer:
Atmosphere to mmHg Conversion Example. Task: Convert 8 atmospheres to mmHg (show work) Formula: atm x 760 = mmHg Calculations: 8 atm x 760 = 6,080 mmHg Result: 8 atm is equal to 6,080 mmHg.
Explanation:
This answer is helpfull for you I nowIf 0.21J of heat cause a 0.308 degree C temperature change, what mass of water is present?
a 0.0702 g
b 0.00540 g
c 0.163 g
d 18.4 g
Answer:
The correct answer is Option c (0.163 g).
Explanation:
Given:
Heat energy,
Q = 0.21 J
Specific heat,
c = 4.184 J/g°c
Change in temperature,
ΔT = 0.308°C
As we know,
⇒ [tex]Q=mc \Delta T[/tex]
By substituting the values, we get
[tex]0.21=m\times 4.184\times 0.308[/tex]
[tex]m=\frac{0.21}{0.308\times 4.184}[/tex]
[tex]=\frac{0.21}{1.28867}[/tex]
[tex]=0.163 \ g[/tex]
Suppose an electron is transferred from a potassium atom to an unknown halogen atom. For which of the following halogen atoms would this process require the least amount of energy?
A. Cl
B. Br
C. I
Answer:
Cl
Explanation:
Electronegativity is the ability of an electron to attract electrons.
Now, due to the fact that halogens need just one more electron to become stable in their outermost shell, it means all halogens are electronegative.
However, the smaller the atomic number, the bigger the charge density and thus the more electronegative.
Thus, it is the halogen element with the highest atomic number further down the periodic table that will have the least electro negativity and thus require highest amount of energy to attract other electrons.
Thus, since chlorine (Cl) has the least atomic number of 17, then it means that it will be the one that will easily accept the electrons the most from other elements. Therefore the process of transferring electrons from potassium to chlorine will take the least amount of energy.
aromatic compounds aliphatic compounds
Answer:
I hoped it helps you fod blessed:)
Which of the following was NOT explained by Dalton's atomic theory?
ANSWER:
A. the Law of Multiple Proportions
B. the difference between elements and compounds
C.?the difference between isotopes of an element
D. the Law of Conservation of Mass
Answer:
i think 1. law of muliple proportion
Explanation:
please like
A sample of hydrogen nitrate or nitric acid, HNO 3 contains 18.8 x 1022 molecules.
How much mass of nitric acid are in the sample?
Answer:
19.7 g.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize this problem can be solved by using a molecules-moles-mass relationship, starting with the given molecules, using the Avogadro's number and the molar mass of nitric acid (63.01 g/mol):
[tex]18.8x10^{22}molec*\frac{1mol}{6.022x10^{23}molec}* \frac{63.01g}{1mol} \\\\=19.7g[/tex]
Regards!
Explain what happens when water reacts with sodium metal. Support your answer with the relevant
equation.
Sodium metal reacts rapidly with water to form a solution of sodium hydroxide (NaOH) and hydrogen gas (H2). This reaction is exothermic.
Equation:
2Na + 2H²0 --------}- 2NaOH + H²
Please please help me
Suppose that you move from a Suppose that you move from a town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust? town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust?
Answer:
all I can say is town near the ocean atmospheric changes will be cooler, warm, sea breeze, and fresh healthy air. Then when it comes to the mountain lot of change firstly there's a dry air
why do people who do a lot of physical work need more carbohydrate?
Answer:
A person doing physical work needs lots of good carbohydrates to keep their energy levels up and proteins to repair a muscle that might get wear and tear from overexertion. Carbohydrate will help the person work for more extended periods.
Calcium chloride and magnesium sulfate are common drying agents. What type of solvent should be dried with calcium chloride, and what type with magnesium sulfate
Answer: The type of solvent that should be dried with calcium chloride is esters while magnesium sulfate is diethyl ether
Explanation:
Drying agents are mainly hygroscopic substances that has the ability to absorb water on exposure to the atmosphere but not enough to form solutions. They are used in desiccators. Examples of drying agents include:
--> CALCIUM CHLORIDE: This is a compound of calcium that is found in soil water and sea water. It is prepared by the action of dilute hydrochloric acid on calcium trioxocarbonate(IV). The anhydrous salt is used in drying a wide variety of solvent including esters.
--> MAGNESIUM SULFATE: This is a slightly acidic drying agent. It works well in solvents like diethyl ether. It is a fast drying agent because it comes as a fine powder with a large surface area.
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4. What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this
Answer:
e. HCOOH and NaCHOO
Explanation:
For a buffer solution, both an acid and its conjugate base are required.
With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.
Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:
H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.
Answer:
a)10.87
b)9.66
c)9.15
d)7.71
e) 5.56
f) 3.43
Explanation:
tep 1: Data given
Volume of 0.030 M NH3 solution = 30 mL = 0.030 L
Molarity of the HCl solution = 0.025 M
Step 2: Adding 0 mL of HCl
The reaction: NH3 + H2O ⇔ NH4+ + OH-
The initial concentration:
[NH3] = 0.030M [NH4+] = 0M [OH-] = OM
The concentration at the equilibrium:
[NH3] = 0.030 - XM
[NH4+] = [OH-] = XM
Kb = ([NH4+][OH-])/[NH3]
1.8*10^-5 = x² / 0.030-x
1.8*10^-5 = x² / 0.030
x = 7.35 * 10^-4 = [OH-]
pOH = -log [7.35 * 10^-4]
pOH = 3.13
pH = 14-3.13 = 10.87
Step 3: After adding 10 mL of HCl
The reaction:
NH3 + HCl ⇔ NH4+ + Cl-
NH3 + H3O+ ⇔ NH4+ + H2O
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.010 L = 0.00025 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00025 =0.00065 moles
Moles HCl = 0
Moles NH4+ = 0.00025 moles
Concentration at the equilibrium:
[NH3]= 0.00065 moles / 0.040 L = 0.01625M
[NH4+] = 0.00625 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.00625/0.01625)
pOH = 4.34
pH = 9.66
Step 3: Adding 20 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.020 L = 0.00050 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00050 =0.00040 moles
Moles HCl = 0
Moles NH4+ = 0.00050 moles
Concentration at the equilibrium:
[NH3]= 0.00040 moles / 0.050 L = 0.008M
[NH4+] = 0.01 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.01/0.008)
pOH = 4.85
pH = 14 - 4.85 = 9.15
Step 4: Adding 35 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.035 L = 0.000875 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000875 =0.000025 moles
Moles HCl = 0
Moles NH4+ = 0.000875 moles
Concentration at the equilibrium:
[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M
[NH4+] = 0.000875 M / 0.065 L = 0.0135 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.0135/3.85*10^-4)
pOH = 6.29
pH = 14 - 6.29 = 7.71
Step 5: adding 36 mL HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.036 L = 0.0009 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.0009 =0 moles
Moles HCl = 0
Moles NH4+ = 0.0009 moles
[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M
Kw = Ka * Kb
Ka = 10^-14 / 1.8*10^-5
Ka = 5.6 * 10^-10
Ka = [NH3][H3O+] / [NH4+]
Ka =5.6 * 10^-10 = x² / 0.0136
x = 2.76 * 10^-6 = [H3O+]
pH = -log(2.76 * 10^-6)
pH = 5.56
Step 6: Adding 37 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.037 L = 0.000925 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000925 =0 moles
Moles HCl = 0.000025 moles
Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M
pH = -log 3.73*10^-4= 3.43
The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:
a) pH = 10.86
b) pH = 9.66
c) pH = 9.15
d) pH = 7.70
e) pH = 5.56
f) pH = 3.43
Calculating the pH a) 0 mL
Initially, the pH of the solution is given by the dissociation of NH₃ in water.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)
The constant of the above reaction is:
[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex] (2)
At the equilibrium, we have:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)
0.030 M - x x x
[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]
After solving for x and taking the positive value:
x = 7.18x10⁻⁴ = [OH⁻]
Now, we can calculate the pH of the solution as follows:
[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]
Hence, the initial pH is 10.86.
b) 10 mL
After the addition of HCl, the following reaction takes place:
NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)
We can calculate the pH of the solution from the equilibrium reaction (3).
[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)
Finding the number of moles of NH₃ and NH₄⁺
The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:
[tex] n_{b} = n_{i} - n_{HCl} [/tex] (6)
[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]
[tex] n_{a} = n_{HCl} [/tex] (7)
[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are given by:
[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex] (8)
[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex] (9)
Calculating the pHAfter entering the values of Ca and Cb into equation (5) and solving for x, we have:
[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]
x = 4.54x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]
Hence, the pH is 9.66.
c) 20 mLWe can find the pH of the solution from the reaction of equilibrium (3).
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are (eq 8 and 9):
[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]
x = 1.40x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]
So, the pH is 9.15.
d) 35 mLWe can find the pH of the solution from reaction (3).
Calculating the concentrations of NH₃ and NH₄⁺[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]
x = 5.013x10⁻⁷ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]
So, the pH is 7.70.
e) 36 mL Finding the number of moles of NH₃ and NH₄⁺[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]
[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]
Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.
At the equilibrium, we have:
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺
Ca - x x x
[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]
[tex] Ka(Ca - x) - x^{2} = 0 [/tex] (10)
Calculating the acid constant of NH₄⁺
We can find the acid constant as follows:
[tex] Kw = Ka*Kb [/tex]
Where Kw is the constant of water = 10⁻¹⁴
[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]
Calculating the pH
The concentration of NH₄⁺ is:
[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]
After solving the equation (10) for x, we have:
x = 2.78x10⁻⁶ = [H₃O⁺]
Then, the pH is:
[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]
Hence, the pH is 5.56.
f) 37 mLNow, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).
Calculating the concentration of HCl
[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]
Calculating the pH
[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]
Therefore, the pH is 3.43.
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viagnesiumi anu
If I have 100g of Magnesium, how much Magnesium Nitride will I theoretically create?
O 24.3g Mg3 N2
O 138.4g Mg3 N2
415.2g Mg3 N2
O 200g Mg3 N2
Answer:
Theoretical yield is 138.4 g
Explanation:
In the first step we determine the reaction:
3Mg + N₂ → Mg₃N₂
Mass of reactant is 100 g. We assume the nitrogen is in excess, so we work with Mg. We convert mass to moles:
100 g . 1mol/ 24.3g = 4.11 moles of Mg.
Ratio is 3:1. 3 moles of Mg can produce 1 mol of nitride
Our 4.11 moles, may produce (4.11 . 1)/3 = 1.37 moles of Mg₃N₂
We convert mass to moles, to find the theoretical yield:
1.37 mol . 100.9 g/mol = 138.2 g
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Answer:
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Explanation:
Alcohols are poor leaving groups.
To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.
Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.
After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.
17. The density of a population would influence which limiting factor?
O niche
O growth rate
O weather
O space
Answer:
The answer is growth rate
Explanation:
it will help you
During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)
Group of answer choices
It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
Answer: The statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.
Explanation:
A chemical reaction in which heat energy is released is called an exothermic reaction. For exothermic reactions, the value of [tex]\Delta H[/tex] is always negative.
A chemical reaction in which heat energy is absorbed is called an endothermic reaction. For endothermic reaction, the value of [tex]\Delta H[/tex] is always positive.
In endothermic reactions, energy required for breaking the bonds between reactants is less than the energy when products are formed due to which the value of [tex]\Delta H[/tex] remains positive.
Thus, we can conclude that the statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.
It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed. The correct option is B.
The above reaction is endothermic because more energy is produced when new bonds form in the products (H = 920 kJ/mol) than is required to break bonds in the reactants (H = -750 kJ/mol).
In an endothermic process, more energy than is generated during bond creation is absorbed from the environment to dissolve existing bonds. This causes a net absorption of energy, which cools the system.
The reaction takes more energy than it releases, proving its endothermic nature, as seen by the positive difference between the energy needed to dissolve bonds and the energy released during bond formation.
Thus, the correct option is B.
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Your question seems incomplete, the probable complete question is:
During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)
Group of answer choices
A. It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
B. It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
C. It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
D. It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
When taking a measurement with a pH meter, keep the instrument in the _______storage solution or water until it is needed. Rinse the pH meter with
_______deionized water or acetone and gently pat dry. Place the meter in the sample solution, and record the measurement when the pH _______stabilizes or reaches the maximum value
Answer:
storage solution , deionized water, stabilizes
Explanation:
A pH meter is a scientific device or instrument that is used to measure the pH of a given aqueous solution thereby determining the nature of the solution whether it is acidic or basic or neutral.
While using the pH meter or taking the measurement using the pH meter --
it should be kept in a storage solution for effective working.Before using the device, it is rinsed with a deionized water and pat dry.Record the measurements when the pH meter stabilizes.You want to quickly set up a temporary water bath in your lab with a volume of 10.0 L and a temperature of 37.0°C. You only have hot water from your hot water faucet (temperature = 61.0°C) and cold water from your cold water faucet (temperature = 22.0°C). What volume of hot water (in liters) must you mix with cold water to get 10.0 L of 37.0°C water? Assume the specific heat of the water is 4.184 J/g･K and that the water has a density of 1.00 g/mL.
Answer:
Volume of hot water required = 3.85L
Explanation:
Suppose volume of hot Then volume of water required cold water = = x L (10.0-x) L
Heat given by hot water (Q₁)
= mass of hot water x heat capacity of water X AT
= x L * 4.184 * J / g. к x(61.0-37.0) °℃.
And Heat absorbed by cold water (Q₂) = (10.0-x) L x 4.184 J/g*k x(37+0 -220) C
Since energy is consumed, Q₁ = Q2.
i.e. X*l *4.184*J/g*k*24C = (10.0-x)L x 184 5
24 x 15 (10.0-x) = 150. - 15x
x = 150. (24+15) = 3.846
So, volume of hot water required. = 3.85 L
When the temperature of the water increases the water becomes hot.
According to the question the volume of hot water required = 3.85L.
Suppose volume of hot Then the volume of water required cold water is [tex]x L (10.0-x) L[/tex]
All the data are given in the question, which is as follows:-
Heat has given by hot water (Q₁)The formula we are going to use is as follows:-
= mass of hot water x heat capacity of water X AT
= [tex]x L * 4.184 *(61.0-37.0) ^oC[/tex]
The heat absorbed by cold water (Q₂) = [tex](10.0-x) L *4.184 *(37+0 -220) ^oC[/tex]
Since energy is consumed, Q₁ = Q2.
[tex]X*l *4.18424C = (10.0-x)L * 184 524 * 15 (10.0-x) = 150. - 15xx = 150. (24+15) = 3.846[/tex]
Hence, the volume of hot water required is = 3.85 L
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What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example
The question is incomplete, the complete question is:
What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.
Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Explanation:
We first calculate the number of moles of soft drink in a volume of 10 mL
The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)
Taking the concentration of soft drink from the example be = 0.375 M
Volume of solution = 10 mL
Putting values in equation 1, we get:
[tex]0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol[/tex]
Calculating volume of sweetened tea:
Moles of sugar = 0.00375 mol
Molarity of sweetened tea = 0.05 M
Putting values in equation 1, we get:
[tex]0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL[/tex]
Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Name the following compound. Group of answer choices 2-methyl-4-pentyne 4-methyl-3-propyl-1-pentyne 3-isopropyl-1-hexyne 1-nonyne 4-propyl-5-hexyne
The question is incomplete, the complete question is shown in the image attached
Answer:
3-isopropyl-1-hexyne
Explanation:
Organic compounds can be named from the structure of the compound. The name reveals the arrangement of atoms and bonds in the molecule.
If we look at the compound in the question, we will notice that the parent chain contains six carbon atoms, the triple bond is located at position 1 and the isopropyl substituent is attached to carbon 3.
Hence the proper name of the compound becomes, 3-isopropyl-1-hexyne.
What are the equipments needed to determine the density of a liquid in laboratory ?
Answer:
A hydrometer is a special device used to determine the density of liquids.
Explanation:
I hope this helps you. Have a nice day!
Which one of the following reactions is NOT balanced?
2 CO + O2 + 2 CO2
2 SO2 + O2 +2 SO3
2 KNO3 + 10 K 5 K20 + N2
SF4 + 3 H2O → H2SO3 + 4HF
Answer:
co+ o2+ 2co2 is not balanced reaction
Choose the correct statement. A) The cathode is the electrode where the oxidation takes place. B) The cathode is the electrode where the reduction takes place. C) Both oxidation and reduction may take place at the cathode, depending on the cell. D) The cathode is always positive
Question 65 pts
(07.02 MC)
During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)
Group of answer choices
It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
Answer:
It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed
Explanation:
A reaction may be endothermic or exothermic. In an endothermic reaction, energy is absorbed by the process while in an exothermic process energy is given out by the process.
Recall that the enthalpy change of a reaction = enthalpy of products - enthalpy of reactants
Hence, where the energy required to break bonds in the reactants is less than the energy released when the products are formed, the reaction is endothermic.
For an endothermic reaction, the enthalpy change of the reaction is positive.
In this case, enthalpy of reaction = 920 - (-750) = 1670 kJ/mol
g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each of the following reactions: a. The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction between BaCl2 and H2SO4 is . c. The brown precipitate formed in the reaction between NaOH and FeCl3 is . d. The blue precipitate formed in the reaction between CuSO4 and NaOH is .
Answer:
For a: Lead iodide is a yellow precipitate.
For b: Barium sulfate is a white precipitate.
For c: Ferric hydroxide is a brown precipitate.
For d: Copper (II) hydroxide is a blue precipitate.
Explanation:
Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.
For the given options:
For (a):The chemical reaction between KI and lead (II) nitrate follows:
[tex]2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)[/tex]
The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.
For b:The chemical reaction between barium chloride and sulfuric acid follows:
[tex]BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)[/tex]
The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.
For c:The chemical reaction between NaOH and ferric chloride follows:
[tex]3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)[/tex]
The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.
For d:The chemical reaction between NaOH and copper sulfate follows:
[tex]CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4[/tex]
The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.
(a) The yellow precipitate formed in the reaction between KI and Pb(NO3)2 would be PbI2 according to the equation:
[tex]Pb(NO_3)_2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO_3(aq)[/tex]
(b) The white precipitate formed in the reaction between BaCl2 and H2SO4 would be BaSO4 according to the equation:
[tex]BaCl_2 (aq) + H_2SO_4 (aq) ---> BaSO_4 (s) + 2 HCl (aq)[/tex]
(c) The brown precipitate formed in the reaction between NaOH and FeCl3 would be Fe(OH)3 according to the equation:
[tex]FeCl_3 (aq) + NaOH (aq) ---> Fe(OH)_3 (s) + NaCl (aq)[/tex]
(d) The blue precipitate formed in the reaction between CuSO4 and NaOH would be Cu(OH)2 according to the equation:
[tex]CuSO_4(aq) + 2 NaOH (aq) ---> Cu(OH)_2 (s) + Na_2SO_4 (aq)[/tex]
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what class of organic compound is formed when cyclopentanone reacts with ethylamine in the presence of trace acid
The question is incomplete, the complete question is;
What functional group results when cyclopentanone reacts with ethylamine in the presence of trace acid? A) cyanohydrin B) semicarbazone C) imine D) enamine E) oxime
Answer:
imine
Explanation:
An imine is an unsaturated amine. An imine contains the carbon- nitrogen double bond.
Imines are obtained when a carbonyl compound is condensed with NH3 or an amine. The reaction involves several steps in its mechanism.
Since cyclopentanone is a ketone (carbonyl compound) and ethylamine is an amine,in the presence of trace acid, condensation of the two compounds occur to yield an imine
How many ozone molecules can each chlorine atom in the stratosphere destroy
Answer:
100,000
Explanation:
what type of properties change ina physical change? Give an example to support your answer?
pls quick who will give the answer first will get the brainliest
Explanation:
We can observe some physical properties, such as density and color, without changing the physical state of the matter observed. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes. A physical change physical change involves a change in physical properties. Examples of physical properties include melting, transition to a gas, change of strength, change of durability, changes to crystal form, textural change, shape, size, color, volume and density.hope it helps.stay safe healthy and happy.