Answers
Answer:
H^+(aq) + OH^-(aq) —> H2O(l)
Explanation:
We'll begin by writing the balanced equation for the reaction.
2HCl(aq) + Ca(OH)2(aq) —> CaCl2(aq) + 2H2O(l)
Ca(OH)2 is a strong base and will dissociates as follow:
Ca(OH)2(aq) —> Ca^2+(aq) + 2OH^-(aq)
HCl is a strong acid and will dissociates as follow:
HCl(aq) —> H^+(aq) + Cl^-(aq)
Thus, In solution a double displacement reaction occurs as shown below:
2H^+(aq) + 2Cl^-(aq) + Ca^2+(aq) + 2OH^-(aq) —> Ca^2+(aq) + 2Cl^-(aq) + 2H2O(l)
To get the net ionic equation, cancel out Ca^2+ and 2Cl^-
2H^+(aq) + 2OH^-(aq) —> 2H2O(l)
H^+(aq) + OH^-(aq) —> H2O(l)
Related Questions
Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)
Answers
Answer:
66.0 atm
Explanation:
We can calculate the osmotic pressure (π) using the following expression.
[tex]\pi = i \times M \times R \times T[/tex]
where,
i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperatureStep 1: Calculate i
Sodium sulfate completely dissociates according to the following equation.
Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻
Since it produces 3 ions, i = 3.
Step 2: Calculate M
We can calculate the molarity of Na₂SO₄ using the following expression.
[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]
Step 3: Calculate T
We will use the following expression.
K = °C + 273.15
K = 20°C + 273.15 = 293 K
Step 4: Calculate π
[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]
The mass of an object with 500 J of kinetic energy moving with a velocity of 5 m/s is kg.
Answers
Answer:
[tex]m=20kg[/tex]
Explanation:
Hello,
In this case, we define the kinnetic energy as:
[tex]K=\frac{1}{2} m*v^2[/tex]
Thus, for finding the mass we simply solve for it on the previous equation given the kinetic energy and the velocity:
[tex]m=\frac{2*K}{v^2}=\frac{500kg*\frac{m^2}{s^2} }{(5\frac{m}{s})^2} =\frac{500kg*\frac{m^2}{s^2} }{25\frac{m^2}{s^2}}\\\\m=20kg[/tex]
Best regards.
Answer:
The answer is 40 kg
Explanation:
You will this formula below:
m=[tex]\frac{2*\\KE}{v^{2} }[/tex]
Now we know our formula, now we plug in the given numbers:
m=[tex]\frac{2(500J)}{(5m/s)^2}[/tex]
Simplify and we get:
m=40 kg
I hope this was helpful.
A chemistry student weighs out of an unknown solid compound and adds it to of distilled water at . After minutes of stirring, only some of the has dissolved. The student drains off the solution, then washes, dries and weighs the that did not dissolve. It weighs 0.570 kg.
Required:
a. Using the information above, can you calculate the solubility of X?
b. If so, calculate it. Remember to use the correct significant digits and units. .
Answers
Complete Question
A chemistry student weighs out 0.950 kg of an unknown solid compound and adds it to 2.00 L of distilled water at . After minutes of stirring, only some of the has dissolved. The student drains off the solution, then washes, dries and weighs the that did not dissolve. It weighs 0.570 kg.
Required:
a. Using the information above, can you calculate the solubility of X?
b. If so, calculate it. Remember to use the correct significant digits and units. .
Answer:
a
Yes the solubility of X can be calculated this is because the solubility of a substance dissolved in a solution is the amount of that substance that is needed to saturate 1 unit volume of the solvent solution at that given temperature.
And from our question we see that substance X saturated the solvent and there is still remained undissolved substance X
b
The solubility of X is [tex]S = 190 g /L[/tex]
Explanation:
From the question we are told that
The initial mass of the unknown solid is [tex]m_i =0. 950 \ kg[/tex]
The mass of the undissolved substance is [tex]m_u = 0.570 \ kg[/tex]
The volume of the solution is [tex]V =2.00\ L[/tex]
Yes the solubility of X can be calculated this is because the solubility of a substance dissolved in a solution is the amount of that substance that is needed to saturate 1 unit volume of the solvent solution at that given temperature.
And from our question we see that substance X saturated the solvent and there is still remained undissolved substance X
The mass of the substance that dissolved ([tex]m_d[/tex] ) is mathematically represented as
[tex]m_d = m_i - m_u[/tex]
[tex]m_d = 0.95 - 0.570[/tex]
[tex]m_d = 0.38 \ kg = 0.38 *1000 = 380 g[/tex]
The solubility of this substance (X) is mathematically represented as
[tex]S = \frac{m_d}{V}[/tex]
substituting values
[tex]S = \frac{ 380}{2}[/tex]
[tex]S = 190 g /L[/tex]
If 25.8 mL of an AgNO3 solution is needed to precipitate all Cl- ions in a 1570 mg of KCl (forming AgCl), what is the molarity of the AgNO3nsolution?
Answers
Answer:
M=0.816M
Explanation:
Hello,
In this case, we should consider the following reaction:
[tex]AgNO_3+KCl\rightarrow KNO_3+AgCl[/tex]
Thus, by knowing the 1:1 molar ratio of silver nitrate and potassium chloride, we can easily compute the moles of silver nitrate precipitating the 1570 mg of potassium chloride considering its molar mass of 74.5513 g/mol:
[tex]n_{AgNO_3}=1570mgKCl*\frac{1gKCl}{1000mgKCl} *\frac{1molKCl}{74.5513gKCl}*\frac{1molAgNO_3}{1molKCl} \\\\n_{AgNO_3}=0.021molAgNO_3[/tex]
Then, by using the volume of silver nitrate in liters (0.0258 L), we can directly compute the molarity:
[tex]M=\frac{0.021molAgNO_3}{0.0258L}\\ \\M=0.816M[/tex]
Regards.
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank with of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 13. mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is [tex]K_c = 14.39[/tex]
Explanation:
The chemical equation for this decomposition of ammonia is
[tex]2 NH_3[/tex] ↔ [tex]N_2 + 3 H_2[/tex]
The initial concentration of ammonia is mathematically represented a
[tex][NH_3] = \frac{n_1}{V_1} = \frac{29}{75}[/tex]
[tex][NH_3] = 0.387 \ M[/tex]
The initial concentration of nitrogen gas is mathematically represented a
[tex][N_2] = \frac{n_2}{V_2}[/tex]
[tex][N_2] = 0.173 \ M[/tex]
So looking at the equation
Initially (Before reaction)
[tex]NH_3 = 0.387 \ M[/tex]
[tex]N_2 = 0 \ M[/tex]
[tex]H_2 = 0 \ M[/tex]
During reaction(this is gotten from the reaction equation )
[tex]NH_3 = -2 x[/tex](this implies that it losses two moles of concentration )
[tex]N_2 = + x[/tex] (this implies that it gains 1 moles)
[tex]H_2 = +3 x[/tex](this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium
[tex]NH_3 = 0.387 -2x[/tex]
[tex]N_2 = x[/tex]
[tex]H_2 = 3 x[/tex]
Now since
[tex][NH_3] = 0.387 \ M[/tex]
[tex]x= 0.387 \ M[/tex]
[tex]H_2 = 3 * 0.173[/tex]
[tex]H_2 = 0.519 \ M[/tex]
[tex]NH_3 = 0.387 -2(0.173)[/tex]
[tex]NH_3 = 0.041 \ M[/tex]
Now the equilibrium constant is
[tex]K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]
substituting values
[tex]K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}[/tex]
[tex]K_c = 14.39[/tex]
The substances nitrogen monoxide and hydrogen gas react to form nitrogen gas and water. Unbalanced equation: NO (g) + H2 (g) N2 (g) + H2O (l) In one reaction, 76.2 g of H2O is produced. What amount (in mol) of H2 was consumed? What mass (in grams) of N2 is produced?
Answers
Answer:
H2 consumed 4.22 mol
N2 produced 59.107 g
Explanation:
Balanced equation:
2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)
To perform the calculations, the molecular weights of the following compounds must be known:H2O MW = 18.02 g/mol
N2 MW = 28.01 g/mol
To determine the moles of H2O produced, the following formula should be used:
[tex]MW=\frac{mass}{mol}[/tex]
The value of moles is cleared:
[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]
Now, to calculate the grams of N2 consumed, we look at the balanced equation and note that 2 moles of H2 produce 1 mole of N2. Therefore, through said observation, the amount of moles of H2 consumed can be determined.2 mol H2 ⇒ 1 mol N2
4.22 mol H2 ⇒ X
[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]
To calculate the mass of H2 consumed, the molecular weight equation is used again:
[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]
In general,for a gas at a constant volume?
Answers
Answer:
The pressure of a gas is directly proportional to its Kelvin temperature if the volume is kept constant. At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases.
Explanation:
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Answers
Answer:
−153.1 J / (K mol)
Explanation:
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Hg(liq) + Cl2(g) → HgCl2(s)
Given that;
The standard molar entropies of the species at that temperature are:
Sºm (Hg,liq) = 76.02 J / (K mol) ;
Sºm (Cl2,g) = 223.07 J / (K mol) ;
Sºm (HgCl2,s) = 146.0 J / (K mol)
The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]
= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]
= -153.09 J / (K mol)
= or -153.1 J / (K mol)
Hence the answer is −153.1 J / (K mol)
Which of the compounds below are amines?
1. H4C-NH-CH3
2. H3C-NH-C-CH3
H2C-CH3
1 +
H3C-CH2-N-CH3
CH3
N
3. H
4.
Answers
Answer:
1. H4C-NH-CH3
2. H3C-NH-C-CH3
H2C-CH3
1 +
H3C-CH2-N-CH3
CH3
N
3. H
4.
.
.
.
.
.
.
The equilibrium constant for the reaction NO2(g)+NO3(g)→N2O5(g) is 2.1x10-20 , therefore: a. At equilibrium, the concentration of products and reactants is about the same. b. At equilibrium, the concentration of products is greater than the reactants. c. At equilibrium, the concentration of reactants is greater than the products
Answers
Answer: c. At equilibrium, the concentration of reactants is greater than the products
Explanation:
Equilibrium constant for a reaction is the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.
For the reaction:
[tex]NO_2(g)+NO_3(g)\rightleftharpoons N_2O_5(g)[/tex]
Equilibrium constant is given as:
[tex]K_{eq}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
[tex]2.1\times 10^{-20}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
When
a) K > 1, the concentration of products is greater than the concentration of reactants
b) K < 1, the concentration of reactants is greater than the concentration of products
c) K= 1, the reaction is at equilibrium, the concentration of reactants is equal to the concentration of products
Thus as [tex]K_{eq}[/tex] is [tex]2.1\times 10^{-20}[/tex] which is less than 1,
the concentration of reactants is greater than the concentration of products
Ba(OH)2:_______.
A. 1 barium atom, 1 oxygen atom and 1 hydrogen atom.
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
C. 1 barium atom, 2 oxygen atoms and 2 hydrogen atoms.
D. 1 barium atom, 2 oxygen atoms and 1 hydrogen atom.
Answers
Answer: D
Explanation: Expand this (OH)2 you will get 2O, 2H
Hence 1Ba, 2O, 2H
Answer:
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
How many moles of PC15 can be produced from 51.0 g of Cl2 (and excess P4)?
Express your answer to three significant figures and include the appropriate units.
LIT....ITS NOT .227 or .228!!!!
Answers
Answer:
0.287 mole of PCl5.
Explanation:
We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 = 51g
Number of mole of Cl2 =..?
Mole = Mass /Molar Mass
Number of mole of Cl2 = 51/71 = 0.718 mole
Next, we shall write the balanced equation for the reaction. This is given below:
P4 + 10Cl2 → 4PCl5
Finally, we determine the number of mole of PCl5 produced from the reaction as follow:
From the balanced equation above,
10 moles of Cl2 reacted to produce 4 moles of PCl5.
Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.
Therefore, 0.287 mole of PCl5 is produced from the reaction.
Which table represents a relation that is not function?
Please
Answers
Answer:
1
Explanation:
Any relation with a repeated input value is not a function.
Table 1 has the input value 2 listed twice, so does not represent a function.
Compare the conjugate bases of these three acids. Acid 1: hypochlorous acid , HClO Acid 2: phosphoric acid , H3PO4 Acid 3: hydrogen sulfide , HS- What is the formula for the weakest conjugate base ?
Answers
Answer:
The weakest conjugate is HClO-.
Explanation:
As a general rule, the stronger the Bronsted-Lowry acid, the weaker its conjugate base, and vice versa.
Acid 1: HClO is a strong acid, hence its conjugate base would be weak
Acid 2: H3PO4 is a weak acid, hence its conjugate base would be strong
Acid 3: hydrogen sulphide is also a moderately weak acid with a moderately strong conjugate base.
In order of increasing strengths:
HClO < H2S < H3PO4
The osmotic pressure exerted by a solution is equal to the molarity multiplied by the absolute temperature and the gas constant . Suppose the osmotic pressure of a certain solution is measured to be at an absolute temperature o of 312. K. Write an equation that will let you calculate the molarity c of this solution.
Answers
Answer:
Explanation:
From the question, osmotic pressure exerted by a solution is equal to the MOLARITY multiplied by the absolute TEMPERATURE and the GAS CONSTANT r.
Let P = osmotic pressure,
C = molarity, then
T = absolute temperature
r=gas constant
The Osmotic pressure Equation exerted by a solution [tex]P=C*T*r[/tex]
[tex]P=CTr[/tex]
Then it was required in the question to write an equation that will let you calculate the molarity c of this solution, and this equation should contain ONLY symbols
C= molarity of the solution
P=osmotic pressure
r = gas constant
T= absolute temperature
[tex]C=P/(rT)[/tex]
The equation that will let us calculate the molarity c of this solution = [tex]C=P/(rT)[/tex]
What is Keq for the reaction 2HCl(9) = H2(g) + Cl2(g)?
Answers
Answer:
Keq= [(Cl2) (H2)] / (HCl)^2
Explanation:
Equilibrium Constant, Keq, is written as products/reactants.
So it's going to be Keq= [(Cl2) (H2)] / (HCl)^2
What is the conjugate acid in the following equation hbr + H2O yields h30 positive + BR negative
Answers
Answer:
HBr + H2O = H3O+ + Br-
So our conjugate acid is the H3O+ to H2O
Explanation:
A conjugate acid of a base results when the base accepts a proton.
Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.
What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Answers
2. Points
Which of the following is not a characteristic of a transverse mechanical
wave?
A. It travels at less than the speed of light.
B. It involves displacing the medium perpendicular to the motion of
the wave
C. It looks a little bit like a snake.
D. It is also known as a compression wave.
Answers
Answer:
D
Explanation:
Logitudinal waves also known as compression waves.
It involves displacing the medium perpendicular to the motion of the wave is not a characteristic of a transverse mechanical wave. Option B is correct.
What are transverse mechanical waves?A transverse mechanical wave is a disturbance created by it to transfer energy from one point to another. while the proposition happens the particle present within the medium get vibrates.
in a transverse wave, the particle present will vibrate up and down and are perpendicular to the wave's propagation direction. The particles shake in a directional wave in the longitudinal wave propagation.
Therefore, is not a characteristic of a transverse mechanical wave. Option B is correct. It involves displacing the medium perpendicular to the motion of the wave.
Learn more about transverse mechanical waves, here:
https://brainly.com/question/23374194
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Hot coffee in a mug cools over time and the mug warms up. Which describes the energy in this system? The average kinetic energy of the particles in the mug decreases. The average kinetic energy of the particles in the coffee increases. Thermal energy from the mug is transferred to the coffee. Thermal energy from the coffee is transferred to the mug.
Answers
Answer:
d
Explanation:
Answer:
D
Explanation:
Edge 2021
A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-
Answers
Answer:
[tex]M=0.213M[/tex]
Explanation:
Hello,
In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:
[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]
[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]
[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]
We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:
[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]
Finally, we compute the molarity:
[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]
Regards.
(a) show that the pressure exerted by a fluid P (in pascals) is given by P= hdg, where h is the column of the fluid in metres, d is density in kg/m3, and g is the acceleration due to gravity (9.81 m/s2). (Hint: see appendix 2.). (b) The volume of an air bubble that starts at the bottom of a lake at 5.24 degree celsius increases by a factor of 6 as it rises to the surface of water where the temperature is 18.73 degree celsius and the air pressure is 0.973 atm. The density of the lake water is 1.02 g/cm3. Use the equation in (a) to determine the depth of the lake in metres.
Answers
Answer:
56.4 m
Explanation:
volume increases by factor of 6, i.e [tex]\frac{V2}{V1}[/tex] = 6
Initial temperature T1 at bottom of lake = 5.24°C = 278.24 K
Final temperature T2 at top of lake = 18.73°C = 291.73 K
NB to change temperature from °C to K we add 273
Final pressure P2 at the top of the lake = 0.973 atm
Initial pressure P1 at bottom of lake = ?
Using the equation of an ideal gas
[tex]\frac{P1V1}{T1}[/tex] = [tex]\frac{P2V2}{T2}[/tex]
P1 = [tex]\frac{P2V2T1}{V1T2}[/tex] = [tex]\frac{0.973*6*278.24}{291.73}[/tex]
P1 = 5.57 atm
5.57 atm = 5.57 x 101325 = 564380.25 Pa
Density Ρ of lake = 1.02 g/[tex]cm^{3}[/tex] = 1020 kg/[tex]m^{3}[/tex]
acceleration due to gravity g = 9.81 [tex]m/s^{2}[/tex]
Pressure at lake bottom = pgd
where d is the depth of the lake
564380.25 = 1020 x 9.81 x d
d = [tex]\frac{564380.25}{10006.2}[/tex] = 56.4 m
Which process is used to make lime (calcium oxide) from limestone (calcium carbonate)?
Answers
Answer:
Explanation:
Calcium oxide is fromed by the decomopostion of CaCO3 at high temperature.
CaCO3 ------> CaO +CO2
Hope this helps you
When 1.550 gg of liquid hexane (C6H14)(C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 ∘C∘C to 38.13 ∘C∘C. Find ΔErxnΔErxn for the reaction in kJ/molkJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘CkJ/∘C.
Answers
Answer:
ΔErxn[tex]= -3.90*10^3KJ[/tex]
Explanation:
Given from the question
T1 = 25.87∘C
T2= 38.13∘C.
C= 5.73Kj/C
CHECK THE ATTACHMENT FOR DETAILED EXPLATION
Enter your answer in the provided box. To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night
Answers
Answer:
409.0 kg of sodium sulfate decahydrate will produce 4.49×10⁵ kJ
of heat energy.
Explanation:
CHECK THE COMPLETE QUESTION BELOW
To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night? Note that sodium sulfate decahydrate will transfer 354 kJ/mol
EXPLANATION
Here we were asked to calculate the amount of heat will be generated by 409.0 kg of sodium sulfate decahydrate at night assuming there Isa complete reaction and 100% efficiency of heat transfer in the process
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S) is needed here, so it must be firstly calculated.
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S)
( 1*20) + (22.98*2) + (16*14)+ (32*14)= 322.186 g/mol.
Thus 409.0 kg of H₂₀Na₂O₁₄S will have a value which is equivalent to = (409000g)/(322.186 g/mol.)
=1269.453mol of H₂₀Na₂O₁₄S.
But it was stated in the the question that per mole of H₂₀Na₂O₁₄S will transfer 354 kJ heat.
Therefore, 1269.453mol will transfer 1269.453× 354 kJ = 4.49×10⁵ kJ of heat.
Hence, 409.0 kg of sodium sulfate decahydrate will produce
4.49×10⁵ kJ of heat energy.
Take a series of observations to determine if process is spontaneous. Based upon those observations, you will create an activity series, listing the metals in order of their reactivity. Second, you will construct a series of virtual galvanic cells and use those to power a stopwatch. Third, you will determine the standard reduction potential of an unknown metal; comparing its reduction potential to a standard list, you will identify the unknown. Finally, you will create a situation in which the cells are not in the standard condition and measure the cell potential; using the Nernst equation, you will determine the concentration of an unknown solution
Answer the below questions for the portion of the activity in which Sn(s) is placed in AgNO3(aq)
1. Is there a reaction? (circle the correct response) Yes / No
2. How many electrons are transferred 4 electrons
3. Write the balanced redox reaction for the combination of AgNO3(aq) and Sn(s)Sn(s) + Ag+(aq) Sn2+(aq) + Ag(s)
Answers
Answer:
Explanation:
2AgNO₃ + Sn ⇄ Sn( NO₃)₂ + 2Ag
Ag⁺/Ag = .80 V
Sn⁺²/Sn = - .14 V
Hence Ag will be reduced and Sn will be oxidised . Hence the reaction will take place . YES .
2 ) 2 electrons are transferred .
3 )
2Ag⁺ + 2e = 2Ag
Sn = Sn⁺² + 2e
---------------------------
2Ag⁺ + Sn = Sn⁺² + 2Ag .
Constructive interference occurs when the compression of one wave meets
up with the compression of a second wave.
A. True
B. False
Answers
Answer:
True
Explanation:
Write a Lewis structure for each atom or ion. Draw the particle by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons and non-bonding electrons. Show the charge of the atom. Particles: S2-, Mg, Mg+2, P.
Answers
Answer:
The Lewis structure to this question can be described as follows:
Explanation:
Structure of Lewis for [tex]S^{2-}[/tex]:
The maximum number of electrons from valence in [tex]S^{2-}[/tex] is 8 (6 from S as well as 2 from negative change).
The valence electrons in the Lewis structure are placed on four sides of the atom.
Thus the structure of Lewis for [tex]S^{2-}[/tex] is as follows:
[tex]\left[\begin{array}{ccc} &. .&\\: &S&:\\&. .&\end{array}\right] ^{2-}[/tex]
Lewis Mg Structure:
Complete valence electrons are 2 in Mg.
The Lewis structure for Mg, therefore, is as follows:
[tex]\ . \\ Mg\\ \ .[/tex]
The Lewis structure for [tex]Mg^{2+}[/tex]
The maximum valence of electrons [tex]Mg^{2+}[/tex] in is= 0.
Thus, the structure for [tex]Mg^{2+}[/tex] is as follows:
[tex]Mg^{2+}[/tex]
Lewis structure for P :
The maximum number of valence electrons in P is = 5.
Thus, the structure for P is=
[tex]\ \ \ . \\ : P \ : \\[/tex]
Given the equation 2KCIO3(s)=2KCI(s) + 3O2(g). A 3.00-g sample of KCIO3 is decomposed and the oxygen at 24 degrees C and 0.982 atm is collected. What volume of oxygen gas will be collected assuming 100% yield?
Answers
Answer:
0.912 L or 912 mL
Explanation:
M(KClO3) = 122.55 g/mol
3.00 g KClO3 * 1 mol/122.55 g = 3.00/122.55 mol =0.02449 mol
2KCIO3(s)=2KCI(s) + 3O2(g)
from reaction 2 mol 3 mol
given 0.02449 mol x
x = 0.02449*3/2 =0.03673 mol O2
T = 24 + 273.15 = 297.15 K
PV = nRT
V= nRT/P = (0.03673 mol*0.082057 L*atm/K*mol*297.15 K)/0.982 atm =
= 0.912 L or 912 mL
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation
Answers
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.
In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.
In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.
This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.
Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.
Learn more about the Markovnikov rule here:
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