Answer:
false, it is not an example of facilitated diffusion
Answer:
TrueExplanation:
When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.
Q1: A stock solution containing Mn+2 ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution: For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL. For solution B, 10.00 mL of solution A was diluted to 250.0 mL. For solution C, 10.00 mL of solution B was diluted to 500.0 mL. Calculate the concentrations of the stock solution and solutions A, B, and C.
The concentration of a solution is the amount of solute in a solution.
We have from the question that he mass of manganese = 1.584 g
Hence;
Amount of [tex]Mn^2+[/tex] = 1.584 g/55g/mol = 0.0288 moles
Recall that;
Number of moles = concentration * volume
Let the concentration of the solution be C
0.0288 moles = C * 1 L
C = 0.0288 moles/ 1 L
C= 0.0288 mol/L
Hence concentration of stock solution = 0.0288 mol/L
For solution A
From the dilution formula;
C1V1 = C2V2
where;
C1 = initial concentration
C2 = final concentration
V1 = initial volume
V2= final volume
C1 = 0.0288 mol/L
V1 = 50.00 mL
C2 = ?
V2 = 1000.0 mL
C2 = 0.0288 mol/L * 50.00 mL/1000.0 mL
C2 = 0.00144 mol/L
Hence, concentration of solution A is 0.00144 mol/L
For solution B
C1V1 = C2V2
C1 = 0.00144 mol/L
V1 = 10.00 mL
C2 = ?
V2 = 250.0 mL
C2 = 0.00144 mol/L * 10.00 mL/250.0 mL
C2 = 0.0000576 mol/L
Hence, concentration of solution B is 0.0000576 mol/L
For solution C
C1 = 0.0000576 mol/L
V1 = 10.00 mL
C2 = ?
V2 = 500.0 mL
C2 = 0.0000576 mol/L * 10.00 mL/500.0 mL
C2= 0.000001152 mol/L
Hence, concentration of solution C is 0.000001152 mol/L
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For each of the processes, determine whether the entropy of the system is increasing or decreasing.
a. A snowman melts on a spring day.
b. A document goes through a paper shredder.
c. A water bottle cools down in a refrigerator.
d. Silver tarnishes
e. Dissolved sugar precipitates out of water to form rock candy.
Explanation:
Entropy refers to the degree of disorderliness of a system.
a. A snowman melts on a spring day.
Entropy is increasing because there is a change in state of matter from solid to liquid. Liquid particles have more freedom f movement compared to solids.
b. A document goes through a paper shredder.
Entropy increases because random, disorganized bits of paper are left.
c. A water bottle cools down in a refrigerator.
Entropy decreases because temperature is directly proportional to entropy.
d. Silver tarnishes
Entropy increases because random bits of the sliver particles are formed.
e. Dissolved sugar precipitates out of water to form rock candy.
Entropy decreases because the random dissolved sugar precipitates are ordered into a rock candy.
Write the equations that represent the first and second ionization steps for sulfuric acid (H2SO4) in water.
Answer:
[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}rightarrow[/tex]
Explanation:
Hello,
In this case, given that the sulfuric acid is a diprotic acid (two hydrogen ions) we can identify two ionization reactions, the first one, showing up the dissociation of the first hydrogen to yield hydrogen sulfate ions and the second one, showing up the dissociation of the hydrogen sulfate ions to hydrogen ions and sulfate ions by separated as shown below:
[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}[/tex]
They are have one-sensed arrow, since sulfuric acid is a strong acid.
Regards.
The equations that represent the first and second ionization steps for sulfuric acid in water are H₂SO₄→HSO₄+H⁺ & HSO₄⁻→SO₄⁻+H⁺ respectively.
What is ionization reaction?Ionization reactions are those reactions in which atom or molecule will convert into ion by bearing a positive or negative charge on itself.
In water in the following way ionization of sulphuric acid takes place:
In the first ionization step one hydrogen atom (H⁺) will loose from the sulphuric acid molecule as:H₂SO₄ → HSO₄⁻ + H⁺
In the second ionization step another hydrogen atom will also loose and we get the sulphate ion (SO₄⁻) and one proton (H⁺) as:HSO₄⁻ → SO₄⁻ + H⁺
Hence, two steps are shown above.
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CaCl2 has which bond?
Answer:
CaCl2 has ionic bond because here calcium gives its electron to the chlorine atom and becomes positivetly charged ion.
Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets: How many moles of each of the gaseous products are produced when 20.1 g of pure hydrazine is ignited in the presence of 20.1 g of pure oxygen
Answer:
[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]
Explanation:
Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:
[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]
Now we can balance the reaction:
[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]
In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):
[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]
[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]
In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].
With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]
We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]
I hope it helps!
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step in balancing a redox reaction under basic conditions is to: ______
Explanation:
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step is the last step when balancing a redox reaction under basic condition.
The last step is to cancel all common terms that arises as a result of the formation of the water molecule. Usually, there's need to balance the water molecules in the reactant and product side of the reaction.
Complete the following nuclear equations. I recommend filling in the atomic number below each symbol. Instead of using superscripts in our answers, put each answer in the corresponding box. A). 54Fe 4He --> 2 1n _____
Answer:
gamma ray
Explanation:
Compared to an atom of C-12, an atom of C-14 has
A) fewer protons
C) more neutrons
B) fewer neutrons
D) more protons
Explanation:
Carbon-12 atoms have stable nuclei because of the 1:1 ratio of protons and neutrons.
Carbon-14 atoms have nuclei which are unstable. C-14 atoms will undergo alpha decay and produce atoms of N-14. Carbon-14 dating can be used to determine the age of artifacts which are not more than 50,000 years old.
If the rate of formation (also called rate of production) of compound C is 2M/s in the reaction A --->2C, what is the rate of consumption of A
Answer:
[tex]r_A=-1\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:
[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]
In such a way, solving the rate of consumption of A, we obtain:
[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]
Clearly, such rate is negative which account for consumption process.
Regards.
In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270.
Use the information she obtained to determine the Ka for this acid.
Ka(experiment) = _____
Answer:
Ka = 6.87x10⁻⁵
Explanation:
The equilibrium of benzoic acid in water is:
C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The equilibrium constant, Ka, is:
Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]
The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.
X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.
Concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - X[C₆H₅COO⁻] = X [H₃O⁺] = XpH is defined as -log [H₃O⁺]. As pH = 2.270
2.270 = -log [H₃O⁺]
10^-2.270 = [H₃O⁺]
5.37x10⁻³M = [H₃O⁺] = X.
Replacing, concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M
[C₆H₅COO⁻] = 5.37x10⁻³M
[H₃O⁺] = 5.37x10⁻³M
Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]
Ka = 6.87x10⁻⁵Calculate the percent error in the atomic weight if the mass of a Zn electrode increased by 0.3681g and 6.514x10-3 moles of Zn was produced. You may assume the molar mass of elemental zinc is 65.38 g/mol.
a. -13.5%
b. 13.52%
c. -13.52%
d. 13.5%
Answer:
13.5%
Explanation:
To answer this you need to understand that percentage error can never be negative as we use absolute value in the equation
To get molar/atomic mass = mass/moles
=0.3681g/(6.514×10^-3)moles
=56.509g/moles
percentage error = (65.38-56.509)/65.38 ×100
=13.56% (this is what I'm getting without rounding off your question wasn't specific on how many d.p to use)
Why don't siblings look exactly alike
Answer:
Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.
I need to name an ionic compound containing a transition metal cation and a halogen anion. Below are the rules I should follow to write the correct name for such compound, but one of the options is incorrect: identify and select it.
a. Identify the metal and write its name first
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
c. From the charge of the anion work out the charge of cation as Roman number in parenthesis: specify this charge in the name as a Roman number in parenthesis.
d. Write the number of the anion after the name of the metal
Answer:
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
Explanation:
The keyword in this problem us "transition metal". Transition metals are found between the group 2 and group 3 elements. They have the d sub shells and also exhibit variable oxidation numbers (valency).
Among the options, the incorrect option is option B.
This is because transition metals d not have a fixed oxidation number and they cannot be obtained by looking up the group in the periodic table.
The iconic compounds obtain a transition of metal caution and a halon anon. As per the rules the correct name of the compounds should be written as to identify the incorrect one.
Option B use the ability to check and to work out the charges (oxidation number) of the transition metal as per the group given in the table. The problem with the keyword is transition metal.Learn more about the ionic compound containing a transition metal.
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The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?
Answer:
[tex]Ksp=1.07x10^{-8}[/tex]
Explanation:
Hello,
In this case, the dissociation reaction is:
[tex]PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)[/tex]
For which the equilibrium expression is:
[tex]Ksp=[Pb^{2+}][I^-]^2[/tex]
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
[tex]Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M[/tex]
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
[tex][Pb^{2+}]=1.39x10^{-3}M[/tex]
[tex][I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M[/tex]
Thereby, the solubility product results:
[tex]Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}[/tex]
Regards.
Solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
The dissociation reaction for lead (II) iodide
[tex]\bold {Pb I^2 (s) \leftrightharpoons Pb^2^+ + 2I^- }[/tex]
Solubility product constant at equilibrium.
[tex]\bold {Ksp = [Pb^2^++[I^-]^2}[/tex]
The molar solubility of the substance can be calculated by using the molar mass,
[tex]\bold {s = \dfrac {0.064}{100 mL} \times 461.2 g/mol = 1.39x10^-^3}[/tex]
Molar ratio between between PbI to lead and iodide ions is 1:1 and 1:2 respectively.
Thus Ksp will be,
[tex]\bold {Ksp =(1.39x10^-^3)(2.78x10^-^3 )^2}\\\\\bold {Ksp = 1.07x 10^-^8}[/tex]
Therefore, solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
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15. How many moles of carbon tetrachloride (CCI) is represented by 543.2 g of carbon tetrachloride? The atomic weight of carbon is 12.01
and the atomic weight of chlorine is 35.45.
O A. 11.4 moles
O B.3.53 moles
C. 5.42 moles
D. 8.35x10 moles
Answer:
well, first off. the formula for carbon tetrachloride is CCl4
We need to find the molar mass of carbon tetrachloride
1(Mass of C) + 4(mass of chlorine)
1(12) + 4(35.5)
12 + 142
154 g/mol
Number of moles of CCl3 in 543.2g CCl3
n = given mass / molar mass
n = 543.2/153
n = 3.53 moles
always remember to brainly the questions you find helpful
Answer:
3.53 moles
Explanation:
1. Explain what the police siren sounds like to Jane:
2. Explain what the police siren sounds like to John:
3. Explain why the police siren sounds different between Jane and John:
Answer:
1. the siren has a lower pitch to Jane
2. the siren has a higher pitch to John
3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter
Explanation:
The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.
This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.
What is the change in the freezing point of a solution made by dissolving 14.7 g of
C6H1206 into 150 ml of water? Answer in 2 decimal.
The density of water is 1.00 g/ml and and Kf = 1.86 °C/m
The change in freezing point is 1.02°C
The change in the freezing point (ΔTf) is calculated from the following equation:
ΔTf = i x Kf x m
Given:
i = 1 (Van't Hoff factor, is equal to 1 because the solute is a nonelectrolyte)
Kf = 1.86°C/m (Freezing point constant of the solvent)
m = molality
We have to calculate the molality (m), which is equal to the moles of solute in 1 kg of solvent:
m = moles solute/ kg solvent
The solute is C₆H₁₂O₆. Thus, we calculate the moles of solute by dividing the mass into its molecular weight (Mw):
Mw(C₆H₁₂O) = (12 g/mol C x 6) + ( 1 g/mol H x 12) + (16 g/mol O x 6) = 180 g/mol
moles solute = mass C₆H₁₂O/Mw(C₆H₁₂O) = 14.7 g/(180 g/mol) = 0.082 mol
The solvent is water. Its mass (in kg) is obtained from the volume and the density, as follows:
kg solvent = V x density = 150 mL x 1.00 g/mL x 1 kg/1000 g = 0.15 kg
Now, we calculate the molality (m):
m = moles of solute/kg solvent = 0.082 mol/(0.15 kg) = 0.546 m
Finally, we calculate the change in freezing point:
ΔTf = i x Kf x m = 1 x 1.86°C/m x 0.546 m = 1.02°C
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The direction of the functional group is called?
Explanation:
they are called hydrocarbyls
pls mark me brainliest
Answer:
The first carbon atom that attaches to the functional group is referred to as the alpha carbon.
A sample of gas is held at constant volume. If the number of moles of this sample of gas is doubled and the pressure of this sample of gas is halved, what happens to the absolute temperature of the gas?
Select one
a. The absolute temperature is doubled.
b. The absolute temperature is halved.
c. The absolute temperature is quadrupled.
d. The absolute temperature is quartered.
e. The absolute temperature stays the same.
Answer:
number of moles of gas increases the volume also increases.
A base which can be used to relieve indigestion
Explanation:
Antacids are medications used to manage the symptoms of indigestion and heartburn. Antacids contain active ingredients that are bases. These allow antacids to neutralize any stomach acid which could be causing digestive discomfort.
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5 kJ/mol)
Answer:
The temperature at which the vapor pressure would be 0.350 atm is 201.37°C
Explanation:
The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.
Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.
The equation for Clausius- Clapeyron Equation can be expressed as:
[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]
where ;
[tex]P_1[/tex] is the vapor pressure at temperature 1
[tex]P_ 2[/tex] is the vapor pressure at temperature 2
∆Hvap = enthalpy of vaporization
R = universal gas constant
Given that:
[tex]P_1[/tex] = 1 atm
[tex]P_ 2[/tex] = 0.350 atm
∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol
[tex]T_1[/tex] = 282 °C = (282 + 273) K = 555 K
R = 8.314 J/mol/k
Substituting the above values into the Clausius - Clapeyron equation, we have:
[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]
[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }[/tex]
[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }[/tex]
[tex]\dfrac{1}{T_2} = 0.002108048708[/tex]
[tex]T_2 = \dfrac{1}{0.002108048708}[/tex]
[tex]\mathbf{T_2 }[/tex] = 474.37 K
To °C ; we have [tex]\mathbf{T_2 }[/tex] = (474.37 - 273)°C
[tex]\mathbf{T_2 }[/tex] = 201.37 °C
Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C
The temperature of the liquid at the given vapor pressure is 201.5 ⁰C.
The given parameters;
boiling point temperature, = 282 ⁰Cvapor pressure, P₂ = 0.35 atmenthalpy of vaporization, ∆Hvap = 28.5 kJ/molThe temperature of the liquid will be determined by applying Clausius- Clapeyron Equation;
[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )[/tex]
where;
R is ideal gas constant = 8.314 J/mol.kT₁ is the initial temperature in Kelvin = 282 + 273 = 555 KP₁ is the initial pressure = 1 atm[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )\\\\ln(\frac{0.35}{1} ) = \frac{28.5 \times 10^3}{8.314} (\frac{T_2 - 555}{555T_2} )\\\\-1.049 = 6.176- \frac{3427.95}{T_2} \\\\\frac{3427.95}{T_2} = 6.176 + 1.049\\\\\frac{3427.95}{T_2} = 7.225\\\\T_2 = \frac{3427.95}{7.225} \\\\T_2 = 474.5 \ K\\\\T_2 = 474.5 - 273 = 201.5 \ ^0C[/tex]
Thus, the temperature of the liquid at the given vapor pressure is 201.5 ⁰C.
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A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the cooling
process. What is the final temperature of the water? The specific heat of water is 4.184 J/g.°C.
A. 112°C
B. 58°C
C. 70°C
D. 84°C
E. 27°C
Answer:
B. 58°C
Explanation:
Hello,
In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:
[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]
In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:
[tex]T_2=T_1+\frac{Q}{mCp}[/tex]
Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:
[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]
Thereby, answer is B. 58°C .
Regards.
Write chemical equations to show how the following bases react with water to produce hydroxide ions:
c. Imidazole (C₃H₄N₂), a weak base
d. Sulfite ion (SO₃²⁻), a weak base
Solution:-1
[tex]\boxed{\sf {C_3H_4N_2\atop imidazole}+{H_2O\atop water}\longrightarrow {C_3H_5N_2\atop Imidazolium}+{OH^-}}[/tex]
There can be another
[tex]\boxed{\sf {C_3H_4N_2\atop imidazole}+{H_2O\atop water}\longrightarrow {C_3HN_2\atop Dicyanomethanoide}+{H_3O^+}}[/tex]
Solution:-2:-
[tex]\boxed{\sf {SO_3^{2-}\atop Sulphate\:ion}+{H_2O\atop Water}\longrightarrow {H_2SO_4\atop Sulphuric\:acid}}[/tex]
A piece of metal has a volume of 30.0cm3 and a mass of 252g. What is its density? what metal do you think this is?
Answer:
Explanation:
get density = D = m / V = 0.252 / 0.00003 = 8400 metal will be Cu => bronzeWhich one of the following compounds will NOT be soluble in water?
A) LiNO3
B) PbS
C) NaOH
D) MgCl2
E) K2SO4
A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100. mL of solution. The solubility product constant for sodium carbonate at this temperature is
Answer:
[tex]Ksp=1.2[/tex]
Explanation:
Hello,
In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):
[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]
Next, as its dissociation reaction is:
[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]
And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):
[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]
Best regards.
A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)+H2O(l)↽−−⇀H3O+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.260 M , [H3O+]=4.00×10−4 M , and [A−]=4.00×10−4 M . Calculate the Ka value for the acid HA.
Answer:
Ka = 6.15x10⁻⁷
Explanation:
Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:
HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)
And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:
Ka = [H₃O⁺] [A⁻] / [HA]
You don't take water in the equilibrium beacuse is a pure liquid
Replacing with the concentrations of the problem:
Ka = [H₃O⁺] [A⁻] / [HA]
Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]
Ka = 6.15x10⁻⁷
Notice
10 January 2018
Gift A Tree
This is to inform students of Class XII to assemble in the ground on Saturday, 31 January
2018 at 8:00 in the morning to participate in a Tree Plantation Ceremony being organised
by the Environment Club as a part of the Farewell Programme.
Amita/Amit
Secretary, Environment Club)
Exercises
As the Head Boy/ Head Girl, Central Public School, draft a notice informing all students about
a wrist watch that was found near the school canteen.answar
Answer:
gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata
Explanation:
vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs
A chemist prepares a solution of silver(I) nitrate(AgNO3) by measuring out 269. mu mol of silver(I) nitrate into a 300. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mmol/L of the chemist's silver(I) nitrate solution.
Answer:
concentration in mmol/L = 8.97 × 10⁻¹ mmol/L
Explanation:
Given that:
the number of moles of silver(I) nitrate(AgNO3) the chemist used in preparing a solution = 269 mmol = 269 × 10⁻³ mmol
The volume of the volumetric flask = 300 mL = 300 × 10⁻³ L
In order to calculate the concentration in mmol/L of the chemist's silver(I) nitrate (AgNO3) solution , we used the formula which can be expressed as;
[tex]concentration \ in \ mmol/L = \dfrac{ number \ of \ mmol}{vol. \ of \ the \ solution}[/tex]
[tex]concentration \ in \ mmol/L = \dfrac{ 269 * 10^{-3 } \ mmol }{300 * 10^{-3} \ L }[/tex]
concentration in mmol/L = 0.8966 mmol/L
concentration in mmol/L = 8.97 × 10⁻¹ mmol/L
In the reaction A + B + C + D, what are the reactants?
O A. Just B
B. Cand D
O c. A and B
O D. A and C
Answer:
C.
Explanation:
I believe that it should be A and B.