Answer:
Not necessarily required, the water bath is used to give energy for reaction and it does not invlove in the reaction. So just normal water can be used in water bath.
Explanation:
a) What is the common-ion effect?b) Given an example of a salt that can decrease the ionization of HNO2 in solution. List all of the substances (species) that would be present in the solution. (Consider the HNO2, the salt, and the water.)
Answer:
Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.
NaNO2
Explanation:
In order to understand exactly what common ion effect is, let us consider a simple unambiguous example. Assuming I have a solution of an ionic substance that contains a cation A and an anion B, this ionic substance has chemical formula AB. Secondly, I have another ionic distance with cation C and anion B, its chemical formula is CB. Both CB and AB are soluble in water to a certain degree as shown by their respective KSp.
If I dissolve AB in water and form a solution, subsequently, I add solid CB to this solution, the solubility of CB in this solution is found to be lees than the solubility of CB in pure water because of the ion B^- which is common to both substances in solution. We refer to the phenomenon described above as common ion effect.
Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.
If I try to dissolve NaNO2 in a solution of HNO2, the solubility of NaNO2 in the HNO2 solution will be less than its solubility in pure water due to common ion effect. Also, the extent of ionization of HNO2 in a system that already contains NaNO2 will be decreased compared to its extent ionization in pure water. This system described here will contain HNO2, water and NaNO2
You are titrating 25.025.0 mL of 0.0100 M Sn2 Sn2 in 1 M HCl HCl with 0.0500 M Tl3 Tl3 resulting in the formation of Sn4 Sn4 and Tl Tl . A PtPt indicator electrode and a saturated Ag|AgCl Ag|AgCl reference electrode are used to monitor the titration. What is the balanced titration reaction
Answer:
Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺
Explanation:
The Sn²⁺ is oxidized to Sn⁴⁺. Whereas Tl³⁺ is reduced to Tl⁺. The half-reactions are:
Sn²⁺ → 2e⁻ + Sn⁴⁺ (Oxidation, loosing electrons)
Tl³⁺ + 2e⁻ →Tl⁺ (Reduction, gaining electrons)
The sum of the reactions gives:
Sn²⁺ + Tl³⁺ + 2e⁻ → 2e⁻ + Sn⁴⁺ + Tl⁺
Subtracting the electrons in both sides of the reaction:
Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺In a zero order reaction, it takes 342 seconds for 75% of a hypothetical reactant to decompose. Determine the half-life t_{1/2} in units of seconds. Do not enter units with your numerical answer. Numeric Answer:
Answer:
228 s
Explanation:
In a zero order reaction, the formula for the half life is given as;
t1/2 = [A]o / 2k
To obtain the rate constant k, we have to use;
[A] = [A]o - kt
kt = [A]o - [A]
From the question;
it takes 342 seconds for 75% of a hypothetical reactant to decompose.
We have;
t = 324
[A] = 25
[A]o = 100
Upon solving for k we have;
kt = [A]o - [A]
k = ( [A]o - [A] ) / t
k = (100 - 25 ) / 342
k = 75 / 342 = 0.2193
Solving for t1/2;
t1/2 = [A]o / 2k
t1/2 = 100 / 2(0.2193)
t1/2 = 100 / 0.4386 = 228 s
An electron in a 3s3s orbital penetrates into the region occupied by core electrons more than electrons in a 3p3p orbital. An electron in a orbital penetrates into the region occupied by core electrons more than electrons in a orbital. true false
Answer:
True
Explanation:
The penetrating ability of electrons in the orbitals is in the order s > p > d > f
An electron in a 3s orbital is closer to the nucleus than the one in a 3p orbital and as a result, there will be lesser shielding effect on it. This low shielding effect experienced by the 3s electron gives it a high penetration ability and hence will be able to easily penetrate regions occupied by core electrons. Conversely, the 3p orbital is farther away from the nucleus, electrons revolving around it are highly shielded which limits their ability to penetrate regions of core electrons.
Note that the maximum electrons that the s orbital can accommodate is 2 while p orbital can accommodate a maximum of 8.
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 525 mL of a solution that has a concentration of Na ions of 1.10 M
Answer:
31.652g of Na3PO4
Explanation:
We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:
Na3PO4 will dessicate in solution as follow:
Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)
From the balanced equation above,
1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.
Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e
xM Na3PO4 = (1.10 x 1)/3
xM Na3PO4 = 0.367M
Therefore, the molarity of Na3PO4 is 0.367M.
Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:
Molarity of Na3PO4 = 0.367M
Volume = 525mL = 525/1000 = 0.525L
Mole of Na3PO4 =..?
Molarity = mole /Volume
0.367 = mole /0.525
Cross multiply
Mole of Na3PO4 = 0.367 x 0.525
Mole of Na3PO4 = 0.193 mole.
Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:
Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol
Mole of Na3PO4 = 0.193 mole
Mass of Na3PO4 =.?
Mass = mole x molar mass
Mass of Na3PO4 = 0.193 x 164
Mass of Na3PO4 = 31.652g
Therefore, 31.652g of Na3PO4 is needed to prepare the solution.
A proposed mechanism for the reaction of NO2 and CO is
Step 1: Slow, endothermic:
2 NO2 (g) → NO (g) + NO3 (g)
Step 2: Fast, exothermic:
NO3 (g) + CO (g) → NO2 (g) + CO2 (g)
Overall reaction, exothermic:
NO2 (g) + CO (g) → NO (g) + CO2 (g)
a. Identify each of the following as a reactant, product, or intermediate:
1. NO2(g)
2. CO(g)
3. NO3(g)
4. CO2(g)
5. NO(g)
b. Draw a reaction coordinate diagram for this reaction. Indicate on this drawing the activation energy for each step and the overall enthalpy change.
Answer:
a.
1. NO2(g): Reactant.
2. CO(g): Reactant.
3. NO3(g): Intermediate.
4. CO2(g): Product.
5. NO(g): Product.
b. See attached picture.
Explanation:
Hello,
a. In this case, given the reactions, we can identify each species as:
1. NO2(g): Reactant because it remains at the left side in the overall reaction.
2. CO(g): Reactant because it remains at the left side in the overall reaction.
3. NO3(g): Intermediate because it is a product in the step 1 and a reactant in step 2, for that reason it is not present in the overall reaction.
4. CO2(g): Product because it remains at the right side in the overall reaction.
5. NO(g): Product because it remains at the right side in the overall reaction.
b. In this case, given that the first step is slow and endothermic, it has a high activation energy and the products will have more energy than the reactants, for that reason the final energy is above the initial point. Moreover, since the second step is fast and exothermic, it has a low activation energy and the products will have less energy than the reactants, for that reason, the reaction coordinate diagram is shown on the attached file.
Regards.
Calculate the mass of NaCl required to prepare 360.0 mL of a 0.35 M solution.
Answer:
Explanation:
Molar mass of NaCl = 58.44
Molar mass of NaCl(58.44g) addding to 1000ml give rise to 1M solution always, this can be writtne as below
58.44----- >1000 ml ---> 1M
X gram ----360 ml ---0.35 M
X Gram= 58.44*360*0.35/1000*1 = 7.3 gram
hope you understood the solution
g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass. Must show your work on scratch paper to receive credit.
Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
475 grams of solid calcium oxide reacts with water vapor to form solid calcium hydroxide.Calculate the heat of reaction.
(A) - 221.1 kcal
(B) + 290.8 kcal
C) - 290.8 kcal
(D) + 324.6 kcal
(E) + 221.1 kcal
Answer:
(A) - 221.1 kcal
Explanation:
Based in the reaction:
CaO(s) + H₂O(g) → Ca(OH)₂ ΔH = -109kJ/mol
When 1 mole of CaO reacts per mole of water vapor producing calcium hydroxide there are released -109kJ
475g of CaO (Molar mass CaO: 56.08g/mol) are:
475g CaO × (1mol / 56.08g) = 8.47 moles of CaO
As 1 mole of CaO in reaction release -109kJ, 8.47 moles release:
8.47 mol CaO × (-109 kJ / 1 mol CaO) = -923.2kJ are released
As 1 kCal = 4.184kJ:
-923.2kJ × (1kCal / 4.184kJ) =
-220.7kCal ≈ (A) - 221.1 kcalwhich proess is part of the carbon cycle
Answer:
The key processes in the carbon cycle are: carbon dioxide from the atmosphere is converted into plant material in the biosphere by photosynthesis.
Explanation:
organisms in the biosphere obtain energy by respiration and so release carbon dioxide that was originally trapped by photosynthesis. ... The carbon becomes part of the .
Mathematical relationship between the rate of a chemical reaction and the
concentration of reactants?
Specific Rate Constant
Collision Theory
Rate Law
Activation energy
Answer:
Specific Rate Constant
Explanation:
Mathematical relationship between the rate of a chemical reaction and the
concentration of reactants is Specific Rate Constant.
calculate the energy in joules and calories required to heat 50.0g silver from 106c to 255c.
Answer:
Explanation:
use this fromula
q = m c ∆t
m is mass of silver =50 g
∆t is difference in temperature= 255-106=149
C= specific heat fo silver ( should be mentioned in your question )
Which describes an effect that ocean currents have on short-term climate change? Ocean currents increase the strength of prevailing winds, which can cool the air and land. Ocean currents can carry cold water, which can cool the air and land. Ocean currents increase hurricane activity, which can raise the temperature of the air and land. Ocean currents can carry warm water, which causes hurricane activity and raises the temperature of the air and land.
Answer: B
Ocean currents can carry cold water, which can cool the air and land.
Explanation:
eet ees wat eet ees
plz mark brainliest
Answer:
B is right
Explanation:
In a chemical equation, which symbol should be used to indicate that a substance is in solution? (s)
Answer:
(aq) meaning aqueous solution
Explanation:
hope it helps .
Which diagram represents the bonding pattern of metals?
Answer:
there's no image can't help without it sorry
The compound diborane can be used as a rocket propellant. What is the percent composition of boron in diborane (B2H6)?
Α.22%
Β.25%
C.39%
D.75%
E.78%
Answer:
Explanation:
atomic weight of B= 10.8
atomic weight of h = 1.0
B2H6 = 2*10.81+6*1= 27.62
out of total 27.62 the Boron contrubtion was = 2*10.81= 21.62
For 27.62 B2H6 ---- 21.62 is B
for 100 B2H6 = 21.62*100 /27.62
= ...........%
can you calcualte and the coorect answer ?
Answer:
The correct answer is E. 78%.
[Image Below]
Combustion of hydrocarbons such as nonane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. 1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid nonane into gaseous carbon dioxide and gaseous water. 2. Suppose of nonane are burned in air at a pressure of exactly and a temperature of . Calculate the volume of carbon dioxide gas that is produced. Round your answer to significant digits.
Answer:
Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)
Part 2: Volume of CO₂ produced = 1223.21 L
Note: the complete second part of the question is given below:
2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.
Explanation:
Part 1: Balanced chemical equation
C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)
Part 2: volume of carbon dioxide produced
From the equation of the reaction;
At s.t.p., I mole of C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂
molar mass of C₉H₂₀ = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L
Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.
O.470 Kg of nonane = 470 g of nonane
470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂
Volume of CO₂ gas produced at 1 atm and 17 °C;
Using P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂/P₂T₁
where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K
Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)
Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂
What is an electrolyte solution?
Answer:
an electrolyte is a sustance that produce an electrically conducting solution when dissolved in a polar solvent , such as water
Explanation:
hope it helps
Which of the following aqueous solutions are good buffer systems?
0.31 M ammonium bromide + 0.39 M ammonia
0.31 M nitrous acid + 0.25 M potassium nitrite
0.21 M perchloric acid + 0.21 M potassium perchlorate
0.16 M potassium cyanide + 0.21 M hydrocyanic acid
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite
0.13 M nitrous acid + 0.12 M potassium nitrite
0.15 M potassium hydroxide + 0.22 M potassium bromide
0.23 M hydrobromic acid + 0.20 M potassium bromide
0.34 M calcium iodide + 0.29 M potassium iodide
0.33 M ammonia + 0.30 M sodium hydroxide
0.20 M nitrous acid + 0.18 M potassium nitrite
0.30 M ammonia + 0.34 M ammonium bromide
0.29 M hydrobromic acid + 0.22 M sodium bromide
0.17 M calcium hydroxide + 0.28 M calcium bromide
0.34 M potassium iodide + 0.27 M potassium bromide
Answer:
Answers are in the explanation.
Explanation:
A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Having this in mind:
0.31 M ammonium bromide + 0.39 M ammonia . Is a good buffer system because ammonia is a weak base and its conjugate base, ammonium ion is in the solution.
0.31 M nitrous acid + 0.25 M potassium nitrite . Is a good buffer system because nitrous acid is the weak acid and nitrite ion its conjugate base.
0.21 M perchloric acid + 0.21 M potassium perchlorate . Perchloric acid is a strong acid. Thus, Is not a good buffer system.
0.16 M potassium cyanide + 0.21 M hydrocyanic acid . Hydrocyanic acid is a weak acid and cyanide ion is its conjugate base. Is a good buffer system.
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite . Hypochlorous acid is a weak acid and hypochlorite ion its conjugate base. Is a good buffer system.
0.13 M nitrous acid + 0.12 M potassium nitrite . Is a good buffer system as I explained yet.
0.15 M potassium hydroxide + 0.22 M potassium bromide . Potassium hydroxide is a strong base. Is not a good buffer system.
0.23 M hydrobromic acid + 0.20 M potassium bromide . HBr is a strong acid. Is not a good buffer system.
0.34 M calcium iodide + 0.29 M potassium iodide . CaI and KI are both salts, Is not a good buffer system.
0.33 M ammonia + 0.30 M sodium hydroxide . Ammonia is a weak base but its conjugate base ammonium ion is not in solution. Is not a good buffer system.
0.20 M nitrous acid + 0.18 M potassium nitrite . Is a good buffer system.
0.30 M ammonia + 0.34 M ammonium bromide . Ammonia and ammonium in solution, Good buffer system.
0.29 M hydrobromic acid + 0.22 M sodium bromide . HBr is a strong acid, is not a good buffer system.
0.17 M calcium hydroxide + 0.28 M calcium bromide . CaOH is a strong base, is not a good buffer system.
0.34 M potassium iodide + 0.27 M potassium bromide. KI and KBr are both salts, is not a good buffer system.
Identify each element below, and give the symbols of the other elements in its group.
a. [Ar] 4s23d104p4
b. [Xe] 6s24f145d2
c. [Ar] 4s23d5.
Answer:
Explanation:
the electron configuration is defined as the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It is
used to describe the orbitals of an atom in its ground state
The valence electrons, electrons in the outermost shell, can be used to know the chemical property
a)
Chemical Name of the Element: Selenium
Chemical Symbol: Se
Group it belong in periodic table:6A
Other Element in the same group:tellurium(Te),,sulfur(S)
atomic number = 34
Selenium is a chemical element that has symbol Se It is a nonmetal which is usually classified as metalloid with properties that are intermediate between the elements above and below in the periodic table.
b)Chemical Name of the Element:Hafnium
Chemical Symbol: Hf
Group it belong in periodic table:4B
Other Element in the same group: Titanium( Ti )Rutherfordium
atomic number: 72
Hafnium is a solid at room temperature.
c)Chemical Name of the Element: Manganese
Chemical Symbol:Mg
Group it belong in periodic table:Mn
Other Element in the same group:Bohrium(Bh) ,Technetium(Tc)
Calculate the temperature and state the appropriate phase(s) (solid, liquid, vapor) for each substance: This temperature is: Fahrenheit A. SubstanceMelting Point (K)Boiling Point (K)Phase(s) Oxygen, O254.7590.19 Methane, CH493.15109.10 Water, H2O273.15373.15
Answer:
Explanation:
At 54.75K melting point, Oxygen is in gas (vapour) phase
At 373.15K boiling point, water is in liquid phase.
At 109.10K boiling point methane is in gas (vapour) phase.
Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.
Question:
When 0.500 g of biphenyl (C₁₂H₁₀) undergoes combustion in a bomb calorimeter, the temperature rises from 26.8 °C to 29.5 °C. Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.
Answer;
-4870kJ/mol(3 significant figure)
Which is regarding enzyme inhbition
Explanation:
An enzyme inhibitor is a molecule that binds to an enzyme and decreases its activity. ... Since blocking an enzyme's activity can kill a pathogen or correct a metabolic imbalance, many drugs are enzyme inhibitors. They are also used in pesticides.
how to use VBT to draw the orbital overlapping diagram
Answer:
from the valence elecrtons configuration is the centre atom.atomised the number of elecrtons pair determine the hybridization .
Explanation:
you can read this note to know the ans
Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass of 10.88 g is added and the volume of the water now reads 49.4 mL. What is the density of the metal in g/mL? Multiple Choice 10.9 1.8 0.250 1.6 0.541
Answer:
1.8g
Explanation:
Initial volume = 43.5ml
Final volume = 49.4ml
Mass = 10.88g
Density = ?
Volume = Final volume - initial volume
= 49.4 - 43.5
= 5.9ml
Density = Mass/volume
Density = 10.88/5.9
= 1.8g/ml
When 2,2-dimethylbutane is subjected to free-radical chlorination, ________ distinct monochlorinated products are possible and ________ of these contain asymmetric carbon atoms.
Answer:
See the explanation
Explanation:
In this case, we have to remember that in the monochlorination products we only have to add one "Cl" with this in mind, we can have several options.
a) 1-chloro-2,2-dimethylbutane
b) (R)-3-chloro-2,2-dimethylbutane
c) (S)-3-chloro-2,2-dimethylbutane
d) 1-chloro-3,3-dimethylbutane
Additionally, from these 4 molecules, we will have 2 enantiomers. (R)-3-chloro-2,2-dimethylbutane and (S)-3-chloro-2,2-dimethylbutane.
See figure 1
I hope it helps!
There four distinct monochlorinated products and two of them contain an asymmetric carbon atom.
Free radicals refers to chemical species that posses an odd number of electrons. An asymmetric carbon atom is a carbon atom that is bonded to four different groups.
When 2,2-dimethylbutane is subjected to free-radical chlorination, four distinct monochlorinated products are possible and two of these contain asymmetric carbon atoms.
The four distinct monochlorinated products are;
1-chloro-2,2-dimethylbutane (R)-3-chloro-2,2-dimethylbutane (S)-3-chloro-2,2-dimethylbutane 1-chloro-3,3-dimethylbutane
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Which represents the size of a population?
the birth rate
number of species in an area
the death rate
number of indivduals in a species
Pls hurry
Answer:
The Answer is D
Explanation:
Populations are shown by seeing how many speecies are in the area.
The number of individuals in a species represents the size of a population. hence, the last option is correct.
What is the population size?It is the actual no of individuals in the population.
The density of the population refers to the measurement of population size for per unit area.
So based on this, we can say that The number of individuals in a species represents the size of a population. hence, the last option is correct.
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What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?Cu + 2AgNO3→Cu(NO3)2+ 2Ag
Answer: 323.61 g of [tex]Ag[/tex] will be produced
Explanation:
The given balanced chemical reaction is :
[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]
According to stoichiometry :
2 moles of [tex]AgNO_3[/tex] require 1 mole of [tex]Cu[/tex]
Thus 3.00 moles of [tex]AgNO_3[/tex] will require=[tex]\frac{1}{2}\times 3.00=1.50moles[/tex] of [tex]Cu[/tex]
Thus [tex]AgNO_3[/tex] is the limiting reagent as it limits the formation of product.
As 2 moles of [tex]AgNO_3[/tex] give = 2 moles of [tex]Ag[/tex]
Thus 3.00 moles of [tex]AgNO_3[/tex] give =[tex]\frac{2}{2}\times 3.00=3.00moles[/tex] of [tex]Ag[/tex]
Mass of [tex]Ag=moles\times {\text {Molar mass}}=3.00moles\times 107.87g/mol=323.61g[/tex]
Thus 323.61 g of [tex]Ag[/tex] will be produced from the given moles of both reactants.
Answer:
B). 2KNO3 Is your answer
Explanation:
how many grams of nahco3 (fm84.007) must be added ro 4.00g of k2co3 (fm 138.206) to give a ph of 10.80 in 500ml of water
Answer:
0.804g of NaHCO₃ you must add
Explanation:
pKa of HCO₃⁻/CO₃²⁻ is 10.32.
It is possible to find pH of a buffer by using H-H equation, thus:
pH = pka + log [A⁻] / [HA]
Where [HA] is concentration of acid (HCO₃⁻) and [A⁻] is concentration of conjugate acid (CO₃²⁻).
Moles of CO₃²⁻ = K₂CO₃ are:
4.00g ₓ (1mol / 138.206g) = 0.0289 moles CO₃²⁻
Replacing:
10.80 = 10.32 + log [0.0289] / [HCO₃⁻]
[HCO₃⁻] = 0.009570 moles you need to add to obtain the desire pH
As molar mass of NaHCO₃ is 84.007g/mol, mass of NaHCO₃ is:
0.009570 moles ₓ (84.007g / mol) =
0.804g of NaHCO₃ you must addAnswer:
[tex]0.8237g[/tex] of [tex]NaHCO_3[/tex] must be addedExplanation:
Second dissociation of [tex]H_2CO_3[/tex] is
[tex]HCO_3^-(aq) <-------> CO_3^2-(aq) + H^+[/tex]
[tex]Ka = [CO_3^{2-}] [H+] / [HCO_3^-] \\\\= 4.7*10^{-11}\\\\pKa = -log(4.7 * 10^{-11}) \\\\= 10.33[/tex]
[tex]pH = pKa + \frac{log([A-]}{[HA])}\\\\required pH = 10.80\\\\10.80 = 10.33 + \frac{log([A-]}{[HA])}\\\\\frac{log([A-]}{[HA])} \\\\= 0.47\\\\\frac{[A-]}{[HA]}\\\\ = 2.951[/tex]
[tex][CO_3^{2-}] / [HCO_3^-] = 2.951[/tex]
[tex][HCO_3^-] = [CO_3^{2-}]/2.951[/tex]
moles of [tex]K_2CO_3[/tex] = [tex]\frac{4}{138.206} = 0.028942mol[/tex]
moles of [tex]CO_3^{2-}[/tex] = [tex]0.028942mol[/tex]
[tex][CO_3^{2-}][/tex] = [tex]\frac{0.028942}{500} * 1000 = 0.05788M[/tex]
[tex][HCO_3^-][/tex] required = [tex]\frac{[CO32-]}{2.951} = \frac{0.05788}{2.951} = 0.01961M[/tex]
moles of [tex]HCO_3^-[/tex] required = [tex]\frac{0.01961}{1000} * 500 = 0.009805mol[/tex]
moles [tex]NaHCO_3[/tex] required = [tex]0.009805mol[/tex]
Therefore,
mass of [tex]NaHCO_3[/tex] required [tex]= 0.009805 * 84.007 = 0.8237g[/tex]
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In supersonic flights, molecules break apart and react chemically. which safety features protect the plane?
Answer:
Explanation:
The heat Shield are materials (usually made of metals) protect us from heat by absoring lots of heat and gradually releasing heat by surrounding air cirucaltion
Answer:
Heat shield
Explanation: Most heat shields consist of one or more layers of stamped metal that are shaped into a shield that is designed to wrap around the exhaust manifold. The shield acts as a barrier and heat sink, preventing the heat from the manifold from reaching any of the components under the hood and potentially causing damage.