Answer:
Explanation:
HF acid
F ⁻ conjugate base , name is fluoride ion .
H₂O acid
OH⁻ conjugate base , name hydroxide ion
H₂CO₃ acid ,
HCO₃⁻ conjugate base , name bicarbonate ion .
HSO₄⁻ acid bisulphate ion
SO₄⁻² ion , conjugate base ,name sulphate ion .
Hydrogen,H2,and nitrogen,N2(g), combined to form ammonia,NH3(g):3H2+N2–>2NH3(g) what amount, in moles, of nitrogen will react 18 mols of hydrogen
Answer:
In this reaction, [tex]6\; \rm mol[/tex] of [tex]\rm N_2[/tex] will react with [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex].
Explanation:
In the balanced equation for this reaction, the ratio between the coefficient of [tex]\rm H_2[/tex] and [tex]\rm N_2[/tex] is [tex]3 : 1[/tex]. That is: [tex]n(\mathrm{H_2}) : n(\mathrm{N_2}) = 3:1[/tex]. That's the same as saying that for every one mole of [tex]\rm N_2[/tex] consumed, three moles of [tex]\rm H_2[/tex] will be consumed.
Rewrite this ratio as a fraction:
[tex]\displaystyle \frac{n(\mathrm{H_2})}{n(\mathrm{N_2})} = \frac{3}{1}[/tex].
Take the reciprocal of both sides to obtain:
[tex]\displaystyle \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})} = \frac{1}{3}[/tex].
It is given that [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex] was consumed; in other words, [tex]n(\mathrm{H_2}) = 18\; \rm mol[/tex]. The question is asking for [tex]n(\mathrm{N_2})[/tex], the number of moles of [tex]\rm N_2[/tex] required. Apply this ratio:
[tex]\begin{aligned}n(\mathrm{N_2}) &= n(\mathrm{H_2}) \cdot \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})}\\ &= 18\; \rm mol \times \frac{1}{3} = 6\; \rm mol\end{aligned}[/tex].
Hence the conclusion: in this reaction, it will take (at least) [tex]6\; \rm mol[/tex] of [tex]\rm N_2[/tex] to react with [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex].
As a system becomes less random, its entropy *
decreases
remains the same
increases
Answer:
decreases
Explanation:
The higher the entropy, the more unpredictable and the more random a value is, this is because entropy describes disorder of a system. If things are less random, that means they are predictable, which means they are more ordered.
3. A thin lead apron is used to protect patients from harmful X rays. If the sheet measures 75.0 cm by 55.0 cm by 0.10 cm, and the density of lead is 11.3 g/cm3, what is the mass of the apron in grams?
Answer:
4.67 kg
Explanation:
Given data
Dimensions of the lead sheet: 75.0 cm by 55.0 cm by 0.10 cmDensity of lead: 11.3 g/cm³Step 1: Calculate the volume of the sheet
The volume of the sheet is equal to the product of its dimensions.
[tex]V = 75.0 cm \times 55.0 cm \times 0.10 cm = 413 cm^{3}[/tex]
Step 2: Calculate the mass of the sheet
The density (ρ) is equal to the mass divided by the volume.
[tex]\rho = \frac{m}{V} \\m = \rho \times V = \frac{11.3g}{cm^{3} } \times 413cm^{3} = 4.67 \times 10^{3} g = 4.67 kg[/tex]
Predict what will be observed in experiment below.
Experiment:
Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500mL of hot water (70°C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature.
He prepares batch B by dissolving sugar in 500mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature.
Predicted observation (Choose one).
O It is likely that more rock candy will be formed in batch A.
O It is likely that less rock candy will be formed in batch A.
O It is likely that no rock candy will be formed in either batch.
O I need more information to predict which batch is more likely to form rock candy.
Answer:
Option A: It is likely that more rock candy will be formed in batch A.
Explanation:
The difference between batch A and batch B is that batch A uses temperature to dissolve the sugar, while the dissolution of the sugar in batch B is produced at room temperature.
When he use temperature (hot water) to dissolve the sugar he is increasing the solubility of the sugar in the water, so in batch A we will have more quantity of sugar dissolved than in the batch B. The cooling of the solution at room temperature favors the formation of bigger sugar crystals in the process of crystallization.
From all of the above, the correct predicted observation is the option A: It is likely that more rock candy will be formed in batch A, for the increase of the solubility by the use of hot water. Also, you say that Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization.
I hope it helps you!
If the volume of a spherical ball is 113.04 cubic inches, what is the radius?
3 inches
9 inches
18 inches
27 inches
Answer:
r= 3 inches
Explanation:
V= (4/3)*pi* r^3
113.04= (4/3)*3.14*r^3
113.04*(3/4)= 3.14*r^3
84.78 = 3.14*r^3
84.78/3.14 = r^3
27 = r^3 Take the cubed root of both sides.
r = 3 inches
What system of units is used by only a small number of countries in the world, including the U.S.?
Answer:
The correct answer is the imperial system.
Explanation:
Only three countries in the world, that is, the United States, Myanmar, and Liberia use the imperial system. These include measurements in the form of inches, ounces, Fahrenheit, and feet. In the imperial system, the distances, height, weight, or area measurements are used eventually that traced back to everyday items or parts of the body.
In comparison to other metric systems, the units used in the imperial system are not further differentiated easily into parts of hundreds or thousands, and are thus, regarded of less use in comparison to other metric systems by some. The real follower of the imperial system at present in the world is the United States.
A measurement gave a mass of .34kg. This is the same as 340g
Answer:
Yes it is. 0.34 kg = 340 g (Kilograms and grams)
Explanation:
When we convert from kg to grams we multiply by 1000 because kilo mean 1000.
0.34*1000= 340
When 177. g of alanine (C3H7NO2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9 °C lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2 °C lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.
Answer:
Van't Hoff factor of KBr is 1.63
Explanation:
Freezing point depression due the addition of a solute follows the formula:
ΔT = Kf×m×i
Where ΔT is change in freezing point, Kf is freezing point depression constant of the solvent X, m is molality of the solution (mol / kg) and i is Van't Hoff factor.
Moles of 177g of alanine (Molar mass: 89.09g/mol) are:
177g × (1mol / 89.09g) = 1.99 moles. In 0.800kg:
1.99mol / 0.800kg = 2.49m
Replacing in freezing point depression formula:
5.9°C = Kf×2.49m×1
Alinine has a Van't Hoff factor of 1
The Kf of the solvent is:
2.37 °C/m
Molality of the 177.0g of KBr solution (Molar mass: 119g/mol) is:
177.0g × (1mol / 119g) = 1.487 moles / 0.800kg = 1.859m
And the freezing point depression formula is:
7.2°C = 2.37°C/m×1.859m×i
1.63 = i
Van't Hoff factor of KBr is 1.63
How human health is expected to be affected by climate change?
“IN YOUR OWN WORDS”
The outer surface of a steel gear is to be hardened by increasing its carbon content; the carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at 850 °C (1123 K) for 10 min increases the carbon concentration to 0.90 wt% at a position 1.0 mm below the surface. Estimate the diffusion time required at 650 °C (923 K) to achieve this same concentration also at a 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. D = 1.1 x 10-6 m2 /s and Qd = 87,400 J/mol C diffusion in a-Fe. (8 points)
Answer:
[tex]\mathbf{t_2 = 75.696 \ min}[/tex]
Explanation:
From the question:
The outer surface of a steel gear is to be hardened by increasing its carbon content
Given that :
Diffusion of heat temperature at [tex]T_1[/tex] 850 °C = 1123 K
Diffusion time [tex]t_1[/tex] = 10 min
diffusion after the carbon concentration at a position [tex]x_1[/tex] ( 1.0 mm) below the surface = 0.90 wt%
Preexponential = 1.1 × 10⁻⁶ m²/s
Activation Energy [tex]Q_d[/tex] = 87400 J/mol
We are to determine the time [tex]t_2[/tex] at 650 °C (923 K) to achieve the same diffusion result as at 850 °C (1123 K) for [tex]t_1[/tex] = 10 min
Considering Fick's second law for the condition of Constant surface concentration; we have:
[tex]\frac{Cx-C_0}{C_s-C_0} = 1-erf(\frac{x}{2\sqrt{Dt} } )[/tex] ------ equation (1)
where;
[tex]C_0 =[/tex] concentration of the diffusing solute atom before diffusion
[tex]C_s[/tex] = Constant surface concentration
[tex]C_x[/tex] = Concentration at depth x after time t
[tex]erf(\frac{x}{2\sqrt{Dt} } )[/tex] = Gaussian error function
At some desired specific concentration of solute [tex]C_1[/tex] in an alloy ; the left side of the above equation (1) thus becomes constant ;
i.e [tex]\frac{Cx-C_0}{C_s-C_0} = \mathbf{ constant}[/tex]
Then ; [tex]\frac{x}{2\sqrt{Dt} }[/tex] = constant
[tex]\frac{x^2}{Dt}[/tex] = constant
Dt = constant
Thus; [tex]D_1t_1 = D_2t_2[/tex]
Therefore, the time [tex]t_2[/tex] at 650°C([tex]T_2[/tex] = 923 K) required to produce the same diffusion on result as at 850°C ([tex]T_1[/tex] = 1123 K) for [tex]t_1[/tex] = 10 min is [tex]t_2 = \frac{D_1t_1}{D_2}[/tex]
We need to first determine the Diffusion coefficient at 1123 K and 923 K ( i.e [tex]D_1[/tex] and [tex]D_2[/tex])
At [tex]T_1[/tex] = 1123 K , Diffusion coefficient [tex]D_1[/tex] is calculated by the equation [tex]D_1 = D_0 exp ( - \frac{Q_d}{RT_1})[/tex] (equation from temperature dependence of the diffusion coefficient)
[tex]D_1 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*1123} )[/tex]
[tex]D_1 = 9.462*10^{-11} \ m^2/s[/tex]
[tex]D_2 = D_0 exp ( - \frac{Q_d}{RT_2})[/tex]
[tex]D_2 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*923} )[/tex]
[tex]D_2 = 1.25*10^{-11} m^2/s[/tex]
[tex]t_2 = \frac{ 9.462*10^{-11}*10}{ 1.25*10^{-11} }[/tex]
[tex]\mathbf{t_2 = 75.696 \ min}[/tex]
What distinguishes effusion from diffision? How are these processes similar?
Answer:
In effusion, a substance escapes through a tiny pinhole, or other hole whereas diffusion is the spreading out of a substance within a dispersing medium.
Effusion and diffusion are similar in terms of rates and speeds.
Explanation:
According to effusion, if there is a small hole in a container, a gas will leak from it. So, in this process a substance escapes through a tiny pinhole, or other hole.
According to diffusion, the two solutions will reach the lowest possible concentration of both, if they are combined and not mixed. So, diffusion is the spreading out of a substance within a dispersing medium.
Effusion and diffusion are similar in terms of rates and speeds.
How do you draw structural formulas 2,4-dimethylhexane; 4-methyl-2-pentene; 4-chloro-7-methyl-2-nonyne
Answer:
Structural formulas are the graphical representation of chemical compunds and shows the chemical bonds between the atoms of a molecule.
structural formulas of following compunds is attached below:
2,4-dimethylhexane - C8H18, it will have single bonds between carbon atom and will have methyl group at position 2 and 4 carbon.
4-methyl-2-pentene - C6H12, it will have double bond at 2nd carbon and methyl group at 4th carbon.
4-chloro-7-methyl-2-nonyne - C10H17Cl, it will have triple bond at 2nd position, chloride group at 4th carbon and a methyl group at 7th carbon.
A gas absorbs 21.39 kJ of energy while expanding against 0.276 atm from a volume of 0.0432 L to 1.876 L. What is the energy change of the gas?
Answer:
Q = 21.896kJ
Explanation:
Q = ?
∇U = 21.39kJ
W = ?
W = P∇V
W = P (V2 - V1)
W = 0.276 × (1.876 - 0.04232)
W = 0.276 × 1.83368
W = 0.5060J
Q = ∇U + W
Q = 21.39 + 0.5060
Q = 21.896kJ
The energy change corresponds to the work done by the system
Answer:
[tex]\Delta E=21.34kJ[/tex]
Explanation:
Hello,
In this case, we should apply the first law of thermodynamics to compute the energy change:
[tex]\Delta E=Q-W[/tex]
Thus, with the given volume change we compute the corresponding work in kJ:
[tex]W=P\Delta V=0.276atm*(1.876L-0.0432L)*\frac{101.325kPa}{1atm}*\frac{1m^3}{1000L}=0.0513kJ[/tex]
Then, we compute the energy change:
[tex]\Delta E=21.39kJ-0.0512kJ\\\\\Delta E=21.34kJ[/tex]
Best regards.
What is the formula of the ion formed when tin achieves a stable electron configuration?
A. Sn2-
B. Sn3+
C. Sn4+
D. Sn4-
Answer:
Sn2–
Explanation:
The formula of the ion when tin achieves a stable electron configuration is Sn⁴⁺. Therefore, option (C) is correct.
What are the oxidation states of tin?Tin is a element with the atomic number 50 and the chemical symbol Sn. Tin is present in group 14 as a post-transition metal. Tin has two common oxidation states, +2 and +4. The oxidation state +4 of tin metal is somewhat more stable.
Sn⁴⁺ ion is chemically comparable to both of its neighbor germanium and lead in group 14. The electronic configuration of the tin metal is [Kr] 4d¹⁰5s²5p². It has four electrons in its valence shell so when it losses four electrons it gets configuration with fully filled subshells.
Sn²⁺ is called the stannous ion, while its compound SnCl₂ is known as stannous chloride. Sn⁴⁺ is called the stannic ion, while its compound SnCl⁴ is stannic chloride which is a volatile liquid.
Learn more about tin metal, here:
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A balloon with a volume of 5.0 L is filled with a gas at 760 tore. If the pressure is reduced to 389 torr without a change in temperature. What will be the volume of the balloon?
Answer:
[tex]V_2=9.97L[/tex]
Explanation:
Hello,
In this case, we can apply the Boyle's law in order to understand the pressure-volume relationship as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, we can solve for the resulting volume at the second state as follows:
[tex]V_2=\frac{P_1V_1}{P_2}=\frac{5.0L*760torr}{389tor} \\\\V_2=9.97L[/tex]
Best regards.
A student runs two experiments with a constant-volume "bomb" calorimeter containing 1500.g of water.
First, a 7.500g tablet of benzoic acid C6H5CO2H is put into the "bomb" and burned completely in an excess of oxygen. (Benzoic acid is known to have a heat of combustion of 26.454 kJ/g.) The temperature of the water is observed to rise from 10.00°C to 36.99°C over a time of 13.0 minutes.
Next, 5.260g of ethanol C2H5OH are put into the "bomb" and similarly completely burned in an excess of oxygen. This time the temperature of the water rises from 10.00°C to 28.03°C.
Use this information, to answer the questions below about this reaction:
C2H5OH(l)+ 3O2(g)→ 2CO2(g)+ 3H2O(g)
a. Is this reaction exothermic, endothermic, or neither?
b. If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in the second experiment.
c. Calculate the reaction enthalpy ΔHrxn per mole of CO2
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Since there has been a rise in the reaction temperature, there has been an exothermic reaction.
The amount of heat energy released in the second step has been -132.54 kJ.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
(a) To determine whether the reaction has been exothermic or endothermic, the heat absorbed or released has been calculated.
Since there has been a rise in the temperature of the solution with the combustion, the reaction has been termed as the exothermic reaction.
(b) Amount of heat released in second experiment:
In the bomb calorimeter:
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_w_a_t_e_r[/tex] has been given a:q = mc[tex]\Delta[/tex]T
q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (36.99 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_w_a_t_e_r[/tex] = 169389.24 J.
[tex]\rm q_b_o_m_b\;[/tex] can be given as:q = C [tex]\Delta[/tex]T
q = c (36.99 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] can be given by:q = mass × heat of combustion of benzoic acid
q = 7.5 g × 26.454 kJ/g
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 198405 J
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_b_o_m_b\;[/tex] = - ([tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_w_a_t_e_r[/tex])
[tex]\rm q_b_o_m_b\;[/tex] = - (198405 J + 169389.24 J )
[tex]\rm q_b_o_m_b\;[/tex] = 29015.76 J.
[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c
29015.76 J = 26.99 [tex]\rm ^\circ C[/tex] × c
c of bomb = 1075.05 J/[tex]\rm ^\circ C[/tex].
For the second reaction of combustion of ethanol:
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_w_a_t_e_r[/tex] has been given as:q = mc[tex]\Delta[/tex]T
q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_w_a_t_e_r[/tex] = 113156.28 J.
[tex]\rm q_b_o_m_b\;[/tex] can be given as:q = C [tex]\Delta[/tex]T
q = 1075.05 J/[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_b_o_m_b\;[/tex] = 19383.15 J
Moles of ethanol = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of ethanol = [tex]\rm \dfrac{5.26\;g}{46.07\;g/mol}[/tex]
Moles of ethanol = 0.11417 mol.
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] for ethanol combustion can be given by:q = moles of ethanol × [tex]\Delta[/tex]H of reaction
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417 × [tex]\Delta[/tex]H of reaction
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - ([tex]\rm q_b_o_m_b\;[/tex] + [tex]\rm q_w_a_t_e_r[/tex])
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - (19383.15 J + 113156.28 J)
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132539.43 J
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132.54 kJ.
The amount of heat energy released in the second step has been -132.54 kJ.
(c) The reaction enthalpy per mole can be given as:
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417 mol × [tex]\Delta[/tex]H of reaction
-132.54 kJ = 0.11417 mol × [tex]\Delta[/tex]H of reaction
[tex]\Delta[/tex]H of reaction = -1160.85 kJ/mol.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
Since there has been a rise in the reaction temperature, there has been an exothermic reaction.
The amount of heat energy released in the second step has been -132.54 kJ.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
For more information about the reaction in the bomb calorimeter, refer to the link:
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What is the relationship between the concentration of the hydronium and hydroxide ion and pH in any water solution?
Answer:
The concentration of hydronium ions and the pH value is related by the equation: pH=-Log[H+]
Explanation:
We have two different concepts pH and concentration of hydronium ions. Lets start with the hydronium ion.
An hydronium ion is a species that is produced by an acid:
[tex]HA~->~H^+~+~A^-[/tex]
Aditionally, we can have the production of hydroxide ions. The subtances that have the capacity to produce this ions are called "bases":
[tex]BOH~->~B^+~+~OH^-[/tex]
Now we can continue "pH"
The pH is a scale that indicates if the substance is and acid (higher concentration of [tex]H^+[/tex]), neutral (equal amounts of [tex]H+[/tex] and [tex]OH^-[/tex]) or basic (higher amount of [tex]OH^-[/tex]).
Finally, the "pH" is calculated with the concentration of the hydronium ions (
[tex]H^+[/tex]), the letter "p" is "-Log", therefore:
[tex]pH=-Log[H^+] [/tex]
I hope it helps!
A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temperature rise, assuming it does not melt?
Answer: The temperature rise is [tex]0.53^0C[/tex]
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed by ice = 5280 J
m = mass of ice = 2.40 kg = 2400 g (1kg=1000g)
c = heat capacity of water = [tex]4.18J/g^0C[/tex]
Initial temperature = [tex]T_i[/tex]
Final temperature = [tex]T_f[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=?[/tex]
Putting in the values, we get:
[tex]5280J=2400g\times 4.18J/g^0C\times \Delta T[/tex]
[tex]\Delta T=0.53^0C[/tex]
Thus the temperature rise is [tex]0.53^0C[/tex]
Answer: 0.580 C
Explanation: On Ck-12 I got it right sooo...
If the half life of a first order reaction is 24 days. Calculate the rate constant for the reaction
Answer:
Half-lives of first order reactions
[A]1/2[A]o=12=e−kt1/2.
ln0.5=−kt.
t1/2=ln2k≈0.693k.
Gneiss rock forms from a great amount of pressure and heat underground. What type of rock is it?
Answer:
Gneiss is a foliated metamorphic rock
Explanation:
Gneiss is a high grade metamorphic rock, meaning that it has been subjected to higher temperatures and pressures .
The statements in the tables below are about two different chemical equilibria. The symbols have their usual meaning, for example AG Gibbs free energy of reaction and stands for the equilibrium constant. stands for the standard In each table, there may be one statement that is false because it contradicts the other three statements. If you find a false statement, check the box next to It. Otherwise, check the "no false statements box under the table. statement false? statement false? Ink>0
AH°
R<1
AG'>0
AG'>0
In > AH">TAS
no false statements
Answer:
see explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
The total measure
of angles in a
triangle is 180° the triangle is?
easy question, need help
Answer:
210 amps
Explanation:
B
How does a mixture of benzoic acid and benzaldehyde can be isolated separately by extractions?
Answer:
Dissolve benxioc acid and benzaldehyde in organic solvent. The two compounds are not miscible.with water. Put the two in separating funnel. Then use aqueous sodium bicarbonate to extract. Benzioc acid will be in aqueous layer as benzioate ion. Benzaldehyde remain insoluble and can be isolated.
Explanation:
Extractions are techniques use to separate desired compounds when mixed together. The mixture is brought in contact with solvent in which the sites substance is soluble and other is insoluble. Extractions use imissicible stages to separate substance from another.
What determines the average kinetic energy of the particles in a gas?
A.
the number of collisions
B
the number of particles
C
the size of the particles
D.
the temperature
Answer:
Explanation:
D
A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the ________ present in the buffer solution. The Ka of hydrazoic acid is 2.5 x 10^-5.
Answer:
The pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.
Explanation:
The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:
HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)
The pH of this solution is:
[tex] pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60 [/tex]
When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:
H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)
The number of moles of NaN₃ after the reaction with HCl is:
[tex] \eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles [/tex]
Now, the number of moles of HN₃ is:
[tex] \eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles [/tex]
Then, the pH of the buffer solution after the addition of HCl is:
[tex] pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43 [/tex]
The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.
Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.
I hope it helps you!
Tellurium is a period 5 chalcogen. Selenium is a period 4 chalcogen. If the only factor affecting ionization energies was the nuclear charge, then electrons would be easier to remove (ionize) from Se than Te. Experimentally the opposite is true. It takes 941.0 kJ/mol of energy to ionize the outermost electron from Se while it only takes 869.3 kJ/mol to ionize from Te. A good model should account for this. Quantum mechanical calculations do predict this but require access to sophisticated software, large amounts of computing power and technical expertise. Slater suggested that some simple empirical rules that take into account electron electron repulsion (or shielding) could give a good estimate of the effective nuclear charge (Zeff). The Zeff for the outermost electron in Te is . The Zeff for the outermost electrons in Se is . According to Slater's calculation of effective nuclear charge (does or does not) predict the correct ordering of ionization energies for Se and Te. A better means of rationalizing ionization energies is to include the atomic as follows: [Z subscript e f f end subscript over r squared] . For Te, r = 136 pm and for Se r = 117 pm. This new model (does or does not) predict the correct ordering of first ionization energies for Se and Te.
Answer:
yes
Explanation: took quiz
A solution contains one or more of the following ions: Ag+, Ca2+, and Fe2+. When sodium chloride is added to the solution, no precipitate forms. When sodium sulfate is added to the solution, a white precipitate forms. Then, the precipitate is filtered off and sodium carbonate is added to the remaining solution, and a precipitate forms.
Which of the ions were present in the original solution?
Write net ionic equations for the formation of each of the precipitates observed.
Explanation:
when NaSO4 added, Ca2+ + SO42- = CaSO4
and Calcium sulphate is a white insoluble precipitate
when NaCO3 added, Fe2+ + CO32- = FeCO3
and it is a green precipitate
If I have a 200 L container filled with nitrogen at a pressure of 1.0 atm, how many moles of nitrogen are present at 25 C?
0
Select one:
O a. 0,085 moles
O b. 81.8 moles
O C. 19.3 moles
O d. 8.18 moles
what kind of air pressure would you find in Bridgeport, Connecticut?
Answer:
Warm
Explanation: because it would be less hot air population