Answer:
Al(OH)₃ + OH⁻ → Al(OH)₄⁻
The compound is called hydroxoaluminate.
Explanation:
Aluminiun Hydroxide → Al(OH)₃
NaOH → Sodium hydroxide.
The Al(OH)₃ is an amphoteric compound, while the NaOH is a strong base. When they react, we may think that, fist of all, the base can dissociate: NaOH → Na⁺ + OH⁻
So the Al(OH)₃ will be a Lewis acid, as it can donate a pair of e⁻
Al(OH)₃ + OH⁻ → Al(OH)₄⁻
Why will the conjugate base of a weak acid affect pH? Select the correct answer below: it will react with hydroxide
Answer:
It will react with water
Explanation:
I know this is an older question, but I just wanted to provide the correct answer.
Since we are dealing with a weak base, and the acid is somewhat stronger, it will react with the water molecules to produce hydronium, which will affect the pH of the solution.
The conjugate base of a weak acid affect pH because it will react with hydronium ion.
A weak acid is an acid that does not dissociate completely in water. On the other hand, a strong acid achieves almost 100% dissociation in water.
Acids dissociate in water to yield the hydronium ion and a conjugate base. For instance, the weak acid, acetic acid is dissociated as follows;
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)
We can see that the conjugate base( CH3COO-) could react with the hydronium ions thereby moving the equilibrium position to the left hand side and affecting the pH by decreasing the hydronium ion concentration.
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Missing parts;
Why will the conjugate base of a weak acid affect pH? Select the correct answer below: O it will react with hydroxide O it will react with water O it will react with hydronium O none of the above
At what temperature in K will 0.750 moles of oxygen gas occupy 10.0 L and exert 2.50 atm of pressure
Answer:
406 K.
Explanation:
The following data were obtained from the question:
Number of mole (n) = 0.750 mole
Volume (V) = 10.0 L
Pressure (P) = 2.50 atm
Temperature (T) =.?
Note: Gas constant (R) = 0.0821 atm.L/Kmol
The temperature, T can be obtained by using the ideal gas equation as follow:
PV = nRT
2.5 x 10 = 0.75 x 0.0821 x T
Divide both side by 0.75 x 0.0821
T = (2.5 x 10) /(0.75 x 0.0821 )
T = 406 K.
Therefore, the temperature is 406 K.
Answer: 406 K
Explanation:
We can rewrite the ideal gas law to solve for T:
PV = nRT
T=PV / nR
We are given the following from the problem:
n=0.750 mol P=2.50 atm V=10.0 L
Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:
T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K
GIVING 100 POINTS PLEASE ANSWER SOON When ponds freeze, they freeze at the surface first before freezing at the bottom. Which property of water explains why this might happen? Density Adhesion High boiling point High specific heat
Answer:
[tex]\huge\boxed{Density}[/tex]
Explanation:
This is because of density. Since Ice is less dense than water, the ice even formed inside the pond then starts floating on the surface of water. Because of this density, Ice floats on water and thus the water freezes at the surface.
Answer:
Density
Explanation:
Water is less dense as a solid than as a liquid, so as ponds freeze, the ice floats at the top and the pond freezes from the top-down.
A reaction mechanism has the following proposed elementary steps:Step 1: A → B + CStep 2: A + B → DStep 3: 2 A + D → C + EIf Step 2 is the rate-limiting step, what would the proposed rate law for this mechanism be?
Answer: [tex]Rate=k[A][B][/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
For reactions which takes place in multiple steps are complex reactions and the order is given by the slowest step which is the rate limiting step.
For the given reaction, the rate limiting step is
[tex]A+B\rightarrow D[/tex]
Rate law will be , [tex]Rate=k[A][B][/tex]
Draw two constitutional isomers that share the molecular formula C3H8S. Your structures will have the same molecular formula but will have different connectivities.
Answer:
Two constitutional isomers for the compound C3H8S are shown in the attachment below
Explanation:
Constitutional isomers are isomers that have the same molecular formula but different connectivity. Two constitutional isomers for the compound C3H8S are shown in the attachment below.
As per the concept of structural isomers, the two constitutional isomers that share the molecular formula C₃H₈S are attached in attachment below.
Structural isomers are defined as the isomers in which atoms are completely arranged in a different order but the molecular formula remains the same.
They are the molecules which have same molecular formula but different connectivities of atoms which depend on the order they are put together. An increase in the number of carbon atoms leads to an increase in the structural isomers.
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A 1.00 liter flask initially contained 0.24 mol NO2 at 700 K which decomposed according to the following equation. When equilibrium was achieved, 0.14 mol NO was present. Calculate Kc. 2NO2(g) ↔ 2NO(g) + O2(g)
Answer:
[tex]Kc=0.14[/tex]
Explanation:
Hello,
In this case, for the given reaction, the equilibrium expression is:
[tex]Kc=\frac{[NO]^2[O_2]}{[NO_2]^2}[/tex]
That in terms of the reaction extent [tex]x[/tex] is written as (initial concentration of NO2 is 0.24 M for 0.24 mol in 1.00 L):
[tex]Kc=\frac{(2*x)^2(x)}{(0.24M-2*x)^2}[/tex]
Moreover, since at the equilibrium 0.14 moles of NO are present (a 0.14-M concentration), we can compute the reaction extent as shown below:
[tex][NO]=2*x=0.14M[/tex]
[tex]x=0.14M/2=0.07M[/tex]
In such a way, knowing [tex]x[/tex], we compute Kc as shown below:
[tex]Kc=\frac{(2*0.07)^2(0.07)}{(0.24M-2*0.07)^2}\\\\Kc=0.14[/tex]
Regards.
The kc is 0.14.
Equivalent expression:Since
[tex]Kc = \frac{[NO]^2[O_2]}{[NO_2]^2}\\\\[/tex]
Here the reaction extent x should be written like the initial concentration of NO2 is 0.24 M for 0.24 mol in 1.00 L.
Now
[tex]Kc = \frac{(2\times\ x)^2}{(0.24M - 2\times x)^2}[/tex]
Since at the equilibrium 0.14 moles of NO are presented so the reaction should be
NO = 2*x = 0.14m
x = 0.07
Now kc should be
[tex]= \frac{(2\times 0.07)^2 (0.07)}{(0.24M - 2\times 0.07)^2}[/tex]
= 0.14
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g The electronic structure of which ONE of the following species cannot be adequately described by a single Lewis formula? (In other words, the electronic structure of which one can only be described by drawing two or more resonance structures?) A) C2H4 B) SO3 2– C) SO3 D) C3H8 E) HCN
Answer:
C) SO3
Explanation:
Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.
SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.
Hence, the correct answer is C) SO3
If a sample of C-14 initially contains 1.6 mmol of C-14, how many millimoles will be left after 2250 years
Answer: 1.2 millimoles will be left after 2250 years
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]
b) Amount left after 2250 years
[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]
[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]
[tex]\log\frac{1.6}{a-x}=0.117[/tex]
[tex]\frac{1.6}{a-x}=1.31[/tex]
[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]
Thus 1.2 millimoles will be left after 2250 years
In the Lewis structure of AB4 where B is more electronegative than A. Both are main group elements where A has 8 valence electrons and each B has 7 valence electrons.
Required:
a. What is the total number of valence electrons?
b. How many lone pairs are in the molecule?
Answer:
1. 36
2. Two
Explanation:
The Lewis structure shows the valence electrons present in a compound. Usually the valence electrons are shown as dot structures around the symbol of the elements involved in the compound.
For a compound AB4 where B is more electronegative than A and A has 8 electrons in its valence shell, there will be thirty six valence electrons on the outermost shell of the molecule.
There are six electron pair domains present in the molecule, four bond pairs and two lone pairs. The molecule is in a square planar geometry.
Answer: a- 36 valence electrons
b- 14 lone pairs
Explanation:
Valence is equal to A + 4B = 8 + 4(7)
With 4 bonds between A and the 4 B, that is 36 valence minus 8 electrons in those pairs = 28. 28 is 14 lone pairs.
Given the data: Ag2O(s), = ‑31.1 kJ mol-1, S° = +121.3 J mol-1 K-1 Ag(s), = 0.00 kJ mol-1, S° = +42.55 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = +205.0 J mol-1 K-1 Calculate the temperature at which = 0 for the reaction, Ag2O(s) → 2 Ag(s) + ½ O2(g). Assume that, since the physical states do not change, and are independent of temperature between ‑50.0 °C and 950.0 °C.
Answer:
[tex]T=469.1K\\\\T=195.9\°C[/tex]
Explanation:
Hello,
In this case, for the given decomposition reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):
[tex]\Delta _rH=2\Delta _fH_{Ag}+\frac{1}{2} \Delta _fH_{O_2}-\Delta _fH_{Ag_2O}\\\\\Delta _rH=2*0.00+\frac{1}{2} *0.00-(-31.1)=31.1kJ/mol[/tex]
Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):
[tex]\Delta _rS=2S_{Ag}+\frac{1}{2} S_{O_2}-S_{Ag_2O}\\\\\Delta _rS=2(42.55)+\frac{1}{2} (205.0)-(121.3)=66.3J/(mol*K)[/tex]
Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the unknown temperature, for such Gibbs energy equaling 0, the temperature (in K and °C) turns out:
[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\0=31.1kJ/mol-T(66.3\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\T=\frac{31.1kJ/mol}{0.0663kJ/(mol*K)} =469.1K\\\\T=195.9\°C[/tex]
Which is within the given rank.
Best regards.
Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
Heat required to raise the temperature of ice from -20 °C to 0 °CBeing the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J
Heat required to convert 0 °C ice to 0 °C waterThe heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)
Heat required to raise the temperature of water from 0 °C to 60 °CIn this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
The amount of heat to absorb is 6,261 J.
Calculation for heat:Heat required to raise the temperature of ice from -20 °C to 0 °C.
The formula for specific heat is used to calculate the amount of heat
Q = c * m * ΔT
Where,
Q =heat exchanged by a body,
m= mass of the body
c= specific heat
ΔT= change in temperature
Given:
m= 10 g,
specific heat of the ice= 2.1
ΔT=0 C - (-20 C)= 20 C
On substituting the values:
Q= 10 g*2.1 *20 C
Q=420 J
Heat required to convert 0 °C ice to 0 °C water.
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
Heat required to raise the temperature of water from 0 °C to 60 °C
m= 10 g,
Specific heat of the water= 4.18
ΔT=60 C - (0 C)= 60 C
On substituting:
Q= 10 g*4.18 *60 C
Q=2,508 J
Thus, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
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What is the oxidizing agent in the redox reaction represented by the following cell notation? Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
Answer:
Silver.
Explanation:
Hello,
In this case, for the redox reaction:
[tex]Ni^0(s)+Ag^+(aq)\rightarrow Ni^{2+}+Ag^0(s)[/tex]
We can see the nickel is being oxidized as its oxidation state increases from 0 to 2+ whereas the oxidation state of silver decreases from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
Best regards.
The oxidizing agent in the redox reaction represented by the following cell notation is Silver.
Calculation of the oxidizing agent:The redox reaction is
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
here the nickel is being oxidized since its oxidation state rises from 0 to 2+ while on the other hand, the oxidation state of silver is reduced from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
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Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indium and cadmium. In(s)|
Answer:
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
Explanation:
Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.
We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
Find the pH of these buffer solutions using the information provided: 1L solution containing 80g of lactic acid (MW
Answer:
pH of the solution is 2.0
Explanation:
The lactic acid is a weak acid that is in equilibrium with water as follows:
Lactic acid + H2O ⇄ Lactate + H₃O⁺
And Ka for lactic acid: 1.38x10⁻⁴
Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]
Initial concentration of lactic acid is (MW: 112.06g/mol):
80g * (1mol / 112.06g) / 1L = 0.714M
The equilibrium concentration of the species in the equilibrium are:
[Lactate] = X
[H₃O⁺] = X
[Lactic acid] = 0.714-X
Replacing in Ka expression:
1.38x10⁻⁴ = [X] [X] / [0.714-X]
9.8532x10⁻⁵ - 1.38x10⁻⁴X = X²
9.8532x10⁻⁵ - 1.38x10⁻⁴X - X² = 0
Solving for X:
X = -1.0x10⁻². False solution, there is no negative concentrations
X = 9.86x10⁻³M. Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 9.86x10⁻³M
and pH = -log [H₃O⁺] = -log 9.86x10⁻³M
pH = 2.0
pH of the solution is 2.0Plzzz help and solve this using factor labeling
Answer:
there are 37,8541 liters in 10 gallons
A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is the molality? 4.44 m LiCl 8.70 m LiCl 4.83 m LiCl 4.01 m LiCl
Answer:
mol LiCl = 4.83 m
Explanation:
GIven:
Solution of LiCl in water XLiCl = 0.0800
Mol of water in kg = 55.55 mole
Find:
Molality
Computation:
mole fraction = mol LiCl / (mol water + mol LiCl)
0.0800 = mol LiCl / (55.55 mol + mol LiCl)
0.0800 mol LiCl + 4.444 mol = mol LiCl
mol LiCl - 0.0800 mol LiCl = 4.444 mol
0.92 mol LiCl = 4.444 mol
mol LiCl = 4.83 m
If you had a cup full of methanol and a pool full of methanol, would the mass change?
Answer:
the mass does not change
Explanation:
"What is the difference between the revertible and nonrevertible rII mutants that Benzer generated?"
Why can long chain fatty acids can form micelles in solutions with pH > 7 but are insoluble in pH < 5
Answer:
In basic conditions that is ( pH > 7 ), the equilibrium shifts towards right and produces a lot of (-ve) negatively charged fatty acids which are polar, In water, since they have long hydrophobic hydrocarbon part, this form micelles where the hydrocarbon part remain inside the sphere and -coo- group remain outside the sphere due to H-bonding interaction with water.
At ( pH < 5 ) I.e acidic conditions, the equilibrium shift to the left giving neutral molecules which can not have stronger H-bonding interaction .
So micelles cant form as they become insoluble.
A sample of carbon dioxide gas has a density of g/L at a pressure of 0.889 atm and a temperature of 55.0 °C. Assume ideal behavior,occupies a volume of 686 mL. If the gas is heated at constant pressure until its volume is 913 mL, the temperature of the gas sample will be:_______ ? °C.
Answer:
163.5 °C
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 686 mL.
Initial temperature (T1) = 55 °C.
Final volume (V2) = 913 mL
Initial pressure (P1) = final pressure (P2) = 0.889 atm
Final temperature (T2) =.?
Next, we shall convert celsius temperature to Kelvin temperature.
This can be done as shown below:
Temperature (K) = Temperature (°C) + 273
T(K) = T (°C) + 273
Initial temperature (T1) = 55 °C.
Initial temperature (T1) = 55 °C + 273
Initial temperature (T1) = 328 K
Next, we shall determine the new temperature of the gas.
Since the pressure is constant, we shall determine the new temperature as follow:
V1/T1 = V2 /T2
Initial volume (V1) = 686 mL.
Final volume (V2) = 913 mL
Initial temperature (T1) = 328 K
Final temperature (T2) =.?
V1/T1 = V2 /T2
686/328 = 913/T2
Cross multiply
686 x T2 = 328 x 913
Divide both side by 686
T2 = (328 x 913)/686
T2 = 436.5 K
Finally, we shall convert Kelvin temperature to celsius temperature.
This can be done as shown below:
Temperature (°C) = Temperature (K) – 273
T (°C) = T(K) – 273
T(K) = 436.5 K
T (°C) = 436.5 – 273
T (°C) = 163.5 °C
Therefore, the temperature of the gas sample is 163.5 °C.
A student mixes 1.0 mL of aqueous silver nitrate, AgNO3 (aq), with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube. What will the student observe
Answer:
AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.
Explanation:
If a student mixes 1.0 mL of aqueous silver nitrate AgNO3 (aq) with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube.
The sodium chloride is being acidified with dilute trioxonitrate (V) acid. Then a few drops of silver trioxonitrate(V) is added afterwards. A white precipitate of silver chloride, which dissolves readily in aqueous ammonia indicates the presence of sodium chloride.
The reaction proceeds as follows:
[tex]\mathtt{AgNO_{3(aq)} + NaCl _{(aq)} \to AgCl _{(s)} + NaNO_3_{(aq)}}[/tex]
From the reaction between AgNO3 (aq) and NaCl (aq), AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.
What creation would you make if you had access to any of the chemical elements? can someone answer this for me please.
Answer:
Depends on the chemicals.
Explanation:
The creation of something depends on the chemical we have. If I have chemicals such as acid and base so I can produce salt and water by mixing acid and base. If I have ethanol, vegetable oil and sodium hydroxide which is a catalyst, I can produce biodiesel which can be used in vehicles instead of fossil fuel. If I have sodium element and chlorine gas, I can produce sodium chloride also known as table salt.
What are some geographic features that could be found in the hydrosphere?
Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.
hydro = water
Answer:
Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!
Explanation:
The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)
Consider the compound hydrazine N2H4 (MW = 32.0 amu). It can react with I2 (MW = 253.8 amu) by the following reaction 2 I2 + N2H4 ------------- 4 HI + N2 (a) How many grams of I2 are needed to react with 36.7 g of N2H4? (b) How many grams of HI (MW = 127.9 amu) are produced from the reaction of 115.7 g of N2H4 with excess iodine?
Answer:Cobb
Explanation:What y'all
Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under these conditions is 4.18·J·g−1K−1 . Round your answer to 3 significant digits.
Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
[tex]c=4.18 \frac{J}{g*K}[/tex]m= 1.30 kg= 1,300 g (1 kg=1,000 g)ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.Replacing:
[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]
Q= 64,121.2 J
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
What is advertising used for? Check all that apply. influencing consumer tastes tracking product popularity increasing product awareness promoting company branding gathering data about potential consumers
Answer:
influencing consumer tastes
increasing product awareness
promoting company branding
Explanation:
Advertising is basically a form of communication using creative ideas and communicating benefits of the products. Advertising plays a very crucial role in product business and some of the important uses of advertising are as follows:
Creative advertisements, influence customers or consumers to buy the product.Advertisings involve information regarding the product and so increases product awareness.Advertising on social media platforms, TVs, radio and newspapers, promotes company branding.Hence, the correct options are:
influencing consumer tastesincreasing product awarenesspromoting company branding
Answer:
1,3,4
Explanation:
I took the test
Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
Highest acidity
anilinium ion
aniline
ammonium ion
secondary amine
amide
Lowest acidity
Answer:
anilinium ion > ammonium ion > amide > aniline > secondary amine
Explanation:
Acidity of amine derivatives can derived from their pKa values.
The rule of thumb for acidity with relation to pKa values is that:
As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.
Hence the stronger the acid , the lower pKa value and the weaker the acid , the stronger the pKa value.
So the pKa value for anilinium ion = 4.6
ammonium ion = 9.4
Amide = 15
Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.
The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.
In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.
Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
the amine derivatives ranking is as follows:
anilinium ion > ammonium ion > amide > aniline > secondary amine
1. For the following reaction, 4.86 g of magnesium nitride are mixed with excess water. The reaction yields 7.18 g of magnesium hydroxide.
magnesium nitride(s) + water(1) –> magnesium hydroxide (aq) + ammonia (aq)
What is the ideal yield of magnesium hydroxide?
What is the percent yield for this reaction?
2. For the following reaction, 6.41 g of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 26.2 g of ammonia.
nitrogen(g) + hydrogen(g) –> ammonia(g)
What is the ideal yield of ammonia?
What is the percent yield for this reaction?
3. For the following reaction, 3.79 g of water are mixed with excess chlorine gas. The reaction yields 8.70 g of hydrochloric acid.
chlorine(g) + water(1) –> hydrochloric acid(aq) + chloric acid (HCIO3)(aq)
What is the ideal yield of hydrochloric acid?
What is the percent yield for this reaction?
Answer:
See explanation
Explanation:
1)
Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)
Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles
1 mole of magnesium nitride yields 3 moles of magnesium hydroxide
0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide
Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g
Percent yield= actual yield/ theoretical yield × 100
Percent yield= 7.18/8.398 × 100/1 = 85.5%
2)
N2(g) + 3H2(g) -------> 2NH3(g)
Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.
From the balanced reaction equation;
3 moles of hydrogen gas yields 2 moles of ammonia
3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia
Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g
Percent yield = actual yield/ theoretical yield ×100
Percent yield = 26.2/36.3 ×100
Percent yield = 72.2%
3)
3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)
Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles
Since
3 mole of water yields 5 mole of HCl
0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl
Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g
Percent yield = actual yield/ theoretical yield × 100/1
Percent yield = 8.70/12.775 ×100
Percent yield = 68.1%
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
When the nuclide carbon-14 undergoes beta decay: The name of the product nuclide is . The symbol for the product nuclide is
Answer:
Explanation:
The equation for the decay is given as;
¹⁴₆C --> X + ⁰₋₁e
For conservation of matter, the mass number and atomic number has to be the same in both the reactant and product side f he equation;
Mass number;
14 = x + 0
x = 14
Atomic Number;
6 = x + (-1)
x = 6 + 1 =7
¹⁴₆C --> ¹⁴₇N + ⁰₋₁e
The name of the product nuclide with atomic number of 7 is Nitrogen. The symbol is; ¹⁴₇N
