Answer:
The equilbrium concentration of NH₃(g) is 2.4 M
Explanation:
The balanced reaction is:
4 NH₃(g) + 7 O₂(g) ⇔ 2 N₂O₄(g) + 6 H₂O(g)
By stoichiometry of the reaction, 2 moles of N₂O₄ are formed from 4 moles of NH₃.
Considering that the concentration is [tex]concentration=\frac{number of moles}{volume}[/tex] and with a volume of 1 liter, it is possible to apply the following rule of three: if 2 M of N₂O₄ are formed from 4 M of NH₃, 0.6 M of N₂O₄ from what concentration of NH₃ are formed?
[tex]concentration of NH_{3}=\frac{0.6 M of N_{2}O_{3} *4MofNH_{3} }{2 M of N_{2}O_{3} }[/tex]
concentration of NH₃= 1.2 M
By subtracting the moles of NH3 in equilibrium from the moles of NH₃ initially, you will see how many moles of NH₃ were converted and remain in equilibrium: 3.6 M - 1.2 M= 2.4 M
The equilbrium concentration of NH₃(g) is 2.4 M
what bsic difference is between NMR and MS spectroscopic techniques?
Answer:
The Nuclear magnetic resonance is the process this technique does not use radiation.
The ms is an sensitive technology can be a massive number and small sample of the blood.
Explanation:
The Nuclear magnetic resonance we look at the both side of that coin.
The technique provides that fatty acid composition and various including amino acids.
These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-
(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.
The MS is an extremely sensitive technology using a very small number of the blood.
(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume
MS can be fine mapping metabolic pathways to sign analytical strategy.
A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?
Answer:
Molarity of the sulfuric acid solution is 4.61M
Explanation:
The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:
2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻
That means, 2 moles of base react with 1 mole of sulfuric acid.
If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:
0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄
As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:
0.200 moles H₂SO₄ / 0.0434L = 4.61M
Molarity of the sulfuric acid solution is 4.61MThe equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that
The given question is incomplete. The complete question is given here :
The equilibrium constant for the reaction is [tex]1.1\times 10^6[/tex] M.
[tex]HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)[/tex]
This value indicates that
A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
B. HCN is a stronger acid than HONO
C. The conjugate base of HONO is [tex]ONO^-[/tex]
D. The conjugate acid of CN- is HCN
Answer: A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
Explanation:
Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.
When [tex]K_{p}>1[/tex]; the reaction is product favoured.
When [tex]K_{p};<1[/tex] ; the reaction is reactant favored.
[tex]When K_{p}=1[/tex]; the reaction is in equilibrium.
As, [tex]K_p>>1[/tex], the reaction will be product favoured and as it is a acid base reaction where [tex]HONO[/tex] acts as acid by donating [tex]H^+[/tex] ions and [tex]CN^-[/tex] acts as base by accepting [tex]H^+[/tex]
Thus [tex]HONO[/tex] is a strong acid thus [tex]ONO^-[/tex] will be a weak conjugate base and [tex]CN^-[/tex] is a strong base which has weak [tex]HCN[/tex] conjugate acid.
Thus the high value of K indicates that [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
What is the osmolarity of a 0.20 M solution of KCI?
A) 0.40 Osmol
B) 0.30 Osmol C) 0.20 Osmol D) 0.80 Osmol
E) 0.10 Osmol
Answer:
Osmolarity of solution of KCI = 0.40 osmol
Explanation:
Given:
KCL ⇒ K⁺ + Cl⁻
Find:
Osmolarity of solution of KCI
When M = 0.20 M
Computation:
1 mole of KCL = 2 osmol
1 M of KCl = 2 Osmolarity
So,
Osmolarity of solution of KCI = 2 × 0.20
Osmolarity of solution of KCI = 0.40 osmol
Zn + 2 HCl --> H2 + ZnCl2 If 1.70 g of Zn are reacted, how many grams of ZnCl2 can be created? Show work and process and I will give brainliest
Explanation:
first find the the number of moles of of zinc .
as the number of moles of zinc and ZnCl2 is same we can calculate the mass of ZnCl2.
If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Answer:
1.3 mL
Explanation:
First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:
1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.
It should have 2 significant figures, because 1.2kg has 2 and we are dividing.
The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.
What is volume?Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.
Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.
This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.
Here, it is given that mass of olive oil is 1.2kg.
We know that,
Density of olive oil = 0.917kg/l.
Volume = mass/density
Volume = 1.2/0.917.
Volume = 1.30 lit.
Volume = 1300mL.
Thus, the volume of olive oil will be 1300 mL.
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A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
(1) Before the addition of any hydrobromic acid, the pH is___________.
(2) After adding 12.0 mL of hydrobromic acid, the pH is__________.
(3) At the titration midpoint, the pH is___________.
(4) At the equivalence point, the pH is________.
(5) After adding 45.1 mL of hydrobromic acid, the pH is_________.
Answer:
(1) Before the addition of any HBr, the pH is 12.02
(2) After adding 12.0 mL of HBr, the pH is 10.86
(3) At the titration midpoint, the pH is 10.73
(4) At the equivalence point, the pH is 5.79
(5) After adding 45.1 mL of HBr, the pH is 1.18
Explanation:
First of all, we have a weak base:
0 mL of HBr is added(CH₃)₂NH + H₂O ⇄ (CH₃)₂NH₂⁺ + OH⁻ Kb = 5.4×10⁻⁴
0.289 - x x x
Kb = x² / 0.289-x
Kb . 0.289 - Kbx - x²
1.56×10⁻⁴ - 5.4×10⁻⁴x - x²
After the quadratic equation is solved x = 0.01222 → [OH⁻]
- log [OH⁻] = pOH → 1.91
pH = 12.02 (14 - pOH)
After adding 12 mL of HBrWe determine the mmoles of H⁺, we add:
0.286 M . 12 mL = 3.432 mmol
We determine the mmoles of base⁻, we have
27.9 mL . 0.289 M = 8.0631 mmol
When the base, react to the protons, we have the protonated base plus water (neutralization reaction)
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 3.432 mm -
4.6311 mm 3.432 mm
We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.
[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M
[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M
We have just made a buffer.
pH = pKa + log (CH₃)₂NH / (CH₃)₂NH₂⁺
pH = 10.73 + log (0.116/0.0860) = 10.86
Equivalence pointmmoles of base = mmoles of acid
Let's find out the volume
0.289 M . 27.9 mL = 0.286 M . volume
volume in Eq. point = 28.2 mL
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 8.0631mm -
8.0631 mm
We do not have base and protons, we only have the conjugate acid
We calculate the new concentration:
mmoles of conjugated acid / Total volume (initial + eq. point)
[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL) = 0.144 M
(CH₃)₂NH₂⁺ + H₂O ⇄ (CH₃)₂NH + H₃O⁻ Ka = 1.85×10⁻¹¹
0.144 - x x x
[H₃O⁺] = √ (Ka . 0.144) → 1.63×10⁻⁶ M
pH = - log [H₃O⁺] = 5.79
Titration midpoint (28.2 mL/2)This is the point where we add, the half of acid. (14.1 mL)
This is still a buffer area.
mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)
mmoles of base = 8.0631 mmol - 4.0326 mmol
[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M
[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M
pH = pKa + log (0.096M / 0.096 M)
pH = 10.73 + log 1 = 10.73
Both concentrations are the same, so pH = pKa. This is the maximum buffering capacity.
When we add 45.1 mL of HBrmmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol
mmoles of base = 8.0631 mmoles
This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 12.8986 mm -
- 4.8355 mm
[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)
[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M
- log [H₃O⁺] = pH
- log 0.0662 = 1.18 → pH
This substituent deactivates the benzene ring towards electrophilic substitution but directs the incoming group chiefly to the ortho and para positions.
A) -F
B) -OCH2CH3
C) -CF3
D) -NHCOCH3
E) -NO2
Answer:
F
Explanation:
Halogens may interact with the benzene ring via inductive or resonance effects. Halogens deactivate the benzene ring by inductive effect rather than by resonance effects.
The lone pairs of electrons present on the halogen atoms may be donated to the ring by resonance, but an opposite effect, the inductive pull (-I inductive effect) of the halogen atoms on electrons away from the benzene ring due to the high electro negativity of the halogens leads to a deactivation of the ring towards electrophilic substitution.
Hence inductive electron withdrawal by the halogen atom predominates over electron donation by resonance effect and the benzene ring g is deactivated towards electrophilic substitution at the ortho and para positions.
The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate boiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC
Answer:
Approximately 100 °C.
Explanation:
Hello,
In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:
[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]
We can solve for the temperature as follows:
[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]
Thus, with the proper units, we obtain:
[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]
Hence, answer is approximately 100 °C.
Best regards.
what is radiologist
Radiologists are medical doctors that treat injuries using medical imaging (radiology)
Answer:
a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.
Explanation:
M(8,7) is the midpoint of rs. The coordinates of s are (9,5) what is the coordinates of r
Answer:
Coordinate or r = (7,9).
Explanation:
Data obtained from the question include the following:
Mid point = (8,7)
Coordinate of S = (9,5)
Coordinate of r =...?
We shall determine the coordinate of r as follow:
Let the coordinate of r be (x2, y2)
Mid point = (x1 + x2)/2 , (y1 + y2)/2
Mid point = (8,7)
Coordinate of S = (9,5)
x1 = 9
y1 = 5
x2 =?
y2 =?
The value of x2 can be obtained as follow:
8 = (x1 + x2)/2
8 = (9 + x2)/2
Cross multiply
9 + x2 = 2 × 8
9 + x2 = 16
Collect like terms
x2 = 16 – 9
x2 = 7
The value of y2 can be obtained as follow:
5 = (y1 + y2)/2
7 = (5 + y2)/2
Cross multiply
5 + y2 = 2 × 7
5 + y2 = 14
Collect like terms
y2 = 14 – 5
y2 = 9
Coordinate of r = (x2, y2)
Coordinate or r = (7,9)
Explain the term isomers?
Answer:
Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.
To infer means to do what?
A. reach a conclusion about data
B. make a hypothesis about data
C. decide to collect some more data
D. state something found indirectly from data
Answer:
A
Explanation:
infer means use data to reach conclusion.
Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer coefficients. The reaction that takes place when chlorine gas combines with aqueous potassium bromide. (Use the lowest possible coefficients. Omit states of matter.)
Answer:
[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].
One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.
Explanation:
Formula for each of the speciesStart by finding the formula for each of the compound.
Both chlorine [tex]\rm Cl[/tex] and bromine [tex]\rm Br[/tex] are group 17 elements (halogens.) Each On the other hand, potassium [tex]\rm K[/tex] is a group 1 element (alkaline metal.) EachTherefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between
The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].
Balanced equation for the reactionWrite down the equation using these chemical formulas.
[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.
Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:
[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.
There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.
Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.
In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.
One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].
Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.
One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].
Hence:
[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:
[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:
[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].
Sometimes a nuclide is referenced by the name of the element followed by the:______
a. atomic number
b. mass number
c. electrical charge
d. none of the above
Answer:
The correct option is d
Explanation:
Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.
33. Hydrocarbons that release pleasant odors are called_________
hydrocarbons. (1 point)
Answer:
Aromatic Hydrocarbons
Explanation:
Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.
Answer:
substituted hydrocarbons
Explanation:
i think
Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.
a) ethylbenzene
b) m-Xylene (1,3- dimethylbenzene)
c) 4-Propyl-3-t-butyltoluene
Answer:
Explanation:
a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .
C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O
O is provided by KMnO₄
b ) In this reaction isophthalic acid is formed .
C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂
c)
4-Propyl-3-t-butyltoluene
In this oxidation , three side chains of ring are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .
The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )
suppose you are titrating vinegar, which is an acetic acid solution
Answer:
0.373 M
Explanation:
The balanced equation for the reaction is given below:
HC2H3O2 + NaOH —> NaC2H3O2 + H2O
From the balanced equation above, the following were obtained:
Mole ratio of the acid, HC2H3O2 (nA) = 1
Mole ratio of the base, NaOH (nB) = 1
Next, we shall write out the data obtained from the question. This include:
Volume of base, NaOH (Vb) = 32.17 mL
Molarity of base, NaOH (Mb) = 0.116 M
Volume of acid, HC2H3O2 (Va) = 10 mL
Molarity of acid, HC2H3O2 (Ma) =..?
The molarity of the acid solution can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 10 / 0.116 x 32.17 = 1
Cross multiply
Ma x 10 = 0.116 x 32.17
Divide both side by 10
Ma = (0.116 x 32.17) /10
Ma = 0.373 M
Therefore, the concentration of the acetic acid is 0.373 M.
A mercury manometer is used to measure pressure in the container illustrated. Calculate the pressure exerted by the gas if atmospheric pressure is 751 torr and the distance labeled is 176 mm.
Answer:
Pressure exerted by the gas is 574.85 torr
Explanation:
Atmospheric pressure = 751 torr
but 1 torr = 1 mmHg
therefore,
atmospheric pressure = 751 mmHg
1 mmHg = 133.3 Pa
therefore,
atmospheric pressure = 751 x 133.3 = 100108.3 Pa
distance labeled (tube section with mercury) = 176 mm
the pressure within the tube will be
[tex]P_{tube}[/tex] = ρgh
where ρ is the density of mercury = 13600 kg/m^3
h is the labeled distance = 176 mm = 0.176 m
g is acceleration due to gravity = 9.81 m/s^2
[tex]P_{tube}[/tex] = 13600 x 9.81 x 0.176 = 23481.216 Pa
The general equation for the pressure in the manometer will be
[tex]P_{atm}[/tex] = [tex]P_{tube}[/tex] + [tex]P_{gas}[/tex]
where [tex]P_{atm}[/tex] is the atmospheric pressure
[tex]P_{tube}[/tex] is the pressure within the tube with mercury
[tex]P_{gas}[/tex] is the pressure of the gas
substituting, we have
100108.3 = 23481.216 + [tex]P_{gas}[/tex]
[tex]P_{gas}[/tex] = 100108.3 - 23481.216 = 76627.1 Pa
This pressure can be stated in mmHg as
76627.1 /133.3 = 574.85 mmHg
and also equal to 574.85 torr
How long should you hold the iron on the hair to heat the strand and set the base ?
A) 5 seconds
B) 15 seconds
C) 30 seconds
D) 1 minute
In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 176 kg/h. Determine the percentage of excess air used during this process.
Answer:
37%
Explanation:
From the question, the equation goes does.
C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.
Mair=Mair/Rin
( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.
33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1
= 176/176+8
X= 0.37
0.37 × 100
X= 37%
An atom with 19 protons and 18 neutrons is a(n)
A. Isotope of potassium(K)
B. Standard atom of argon(Ar)
C. Standard atom of (K)
D. Isotope of argon (Ar)
Answer:
A
Explanation:
The number of protons indicates the element so we know it's potassium. To get the number of neutrons you subtract the number of protons (19) from the mass number which for potassium is 39.
39-19=20 neutrons
Because you have 18 neutrons then yours would be an isotope.
Answer: A. Isotope of potassium(K)
Explanation: Founders Educere answer.
Classify each of these reactions.
1) Ba(ClO3)2(s)--->BaCl2(s)+3O2(g)
2) 2NaCl(aq)+K2S(aq)--->Na2S(aq)+2KCl(aq)
3) CaO(s)+CO2(g)--->CaCO3(s)
4) KOH(aq)+AgCl(aq)---->KCl(aq)+AgOH(s)
5) Ba(OH)2(aq)+2HNO2(aq)--->Ba(NO2)2(aq)+2H2O(l)
Each classify reaction should be either one of this.
a. acid-base neutralization
b. precipitation
c. redox
d. none of the above
Answer:
1. REDOX
2. None of the above
3. Precipitation
4. Preicipitation
5. Acid base neutralization
Explanation:
Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.
Option 4 and 3.
3) CaO (s) + CO₂ (g) → CaCO₃(s)
4) KOH (aq) + AgCl (aq) → KCl (aq) + AgOH(s)
Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.
5) Ba(OH)₂ (aq) + 2HNO₂(aq) → Ba(NO₂)₂ (aq) + 2H₂O(l)
Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.
1) Ba(ClO₃)₂ (s) → BaCl₂ (s) + 3O₂(g)
Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).
2. 2NaCl(aq) + K₂S(aq) Na₂S (aq) + 2KCl (aq)
None of the above
Which of the following is a property of salts? Undergo combustion Do not make ionic bonds easily Do not conduct electricity as solids Formed due to reaction of acid with water
Answer:
Do not conduct electricity as solids.
Explanation:
Hello,
In this case, we should remember that salts are formed when an acid and base react in order to yield the salt and water due to the ions exchange during neutralization chemical reactions. For instance, when hydrochloric acid (acid) reacts with potassium hydroxide (base), sodium chloride (salt) and water are yielded via:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
Moreover, it is widely known that salts are formed by electrovalent/ionic bonds which involves electron transfer so the metallic atom becomes positively charged (cation) whereas the non-metallic atom becomes negatively charged (anion) once the electrons are received so it can conduct electricity when dissolved in water yet not when solid since electron transfer is facilitated by the aqueous media, otherwise, ions remain together. Thereby, answer is do not conduct electricity as solids.
Regards.
Answer:
c
Explanation:
A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.
The addition of 0.242 L of 1.92 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions as AgCl and PbCl2. The total mass of the resulting precipitate is 65.08 g. Find the mass of PbCl2 and AgCl in the precipitate. Calculate the mass of PbCl2 and AgCl in grams.
Answer:
Mass PbCl₂ = 50.24g
Mass AgCl = 14.84g
Explanation:
The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:
Ag⁺ + Cl⁻ → AgCl(s)
Pb²⁺ + 2Cl⁻ → PbCl₂(s)
If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:
Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂
And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:
(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl
Moles of Cl⁻ that were added in the KCl solution are:
0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.
Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)
0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles
0.45409 + 0.00021X = 0.46464
0.00021X = 0.01055
X = 0.01055 / 0.00021
X = 50.24g
As X = Mass PbCl₂
Mass PbCl₂ = 50.24gAnd mass of AgCl = 65.08 - 50.24
Mass AgCl = 14.84gThe masses of the compounds in the precipitate can be found my knowing
the number of moles of chloride ion contributed by each compound.
The mass of PbCl₂ in the precipitate is approximately 49.24 gThe mass of AgCl in the precipitate is approximately 15.84 gReasons:
The given parameter are;
Volume of KCl solution added = 0.242 L
Concentration of KCl solution = 1.92 M KCl
The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions
Precipitates formed = AgCl and PbCl₂
The mass of the precipitate = 65.08 g
Required:
The mass of PbCl₂ and AgCl in the precipitate
Solution;
Number of moles of chloride ions in a mole of PbCl₂ = 2 moles
Number of moles of chloride ions in a mole of AgCl = 1 mole
Let X represent the mass of PbCl₂ in the precipitate, we have;
The mass of AgCl in the precipitate = 65.08 g - X
[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]
Number of moles of chloride ions from PbCl₂ is therefore;
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]
The number of moles of chloride ions from one mole of KCl = 1 mole
Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;
0.242 L × 1.92 moles/L = 0.46464 moles
Number of moles of chloride ions from KCl = 0.46464 moles
[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]
Which gives;
[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]
Therefore;
[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]
The mass of PbCl₂ in the precipitate, X ≈ 49.24 g
The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g
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The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) BaSeO4, 0.0118 g/100 mL
(b) Ba(BrO3)2 H20, 0.30 g/100 mL
(c) NH4MgAsO4-6H20, 0.038 g/100 mL
(d) La2(MoOs)3, 0.00179 g/100 mL
Answer:
(a) [tex]Ksp=4.50x10^{-7}[/tex]
(b) [tex]Ksp=1.55x10^{-6}[/tex]
(c) [tex]Ksp=2.27x10^{-12}[/tex]
(d) [tex]Ksp=1.05x10^{-22}[/tex]
Explanation:
Hello,
In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:
(a) [tex]BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}[/tex]
In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}[/tex]
(B) [tex]Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}[/tex]
In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}[/tex]
(C) [tex]NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}[/tex]
In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:
[tex]Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}[/tex]
(D) [tex]La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}[/tex]
In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:
[tex]Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}[/tex]
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Solid MgO has the same crystal structure as NaCl. How many oxide ions surround each Mg * ion as nearest neighbors in MgO? 4 none of these
Answer:
The number of oxide ions as the nearest neighbors of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six
Explanation:
The regularity of a crystal structure leads to the idea of space lattice.In order to explain this concept, let us consider a crystal of NaCl, It consists of a perfectly regular arrangement of sodium ions and chlorine ions.
If we represent the position of each Na+ in the crystal by a point marked x the result will be a regular three dimensional network of points. This will be the space lattice of Na+ in the crystal NaCl. The symmetry of the combined lattice determined the symmetry of the crystal as a whole.
The space lattice of a crystal may be considered as built up of a three dimensional basic pattern called unit cell. The unit cell is a repeat unit which generates the whole pattern in three dimensions of the unit cell.
In Solid MgO , the crystal structure which is used to predict the properties of the material, have the same structure as that of NaCl.
The obtain the structure of a face centered cubic FCC unit cell where the ions occupy the corner of the cube and the center of each face of the cube.
The number of oxide ions as the nearest neighbors of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six. As a result of that , the coordination number of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions is six.
Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accurately the actual electronic structure. Equivalent resonance structures occur when there are identical patterns of bonding within the molecule or ion. The actual structure is a composite, or resonance hybrid, of the equivalent contributing structures. Draw Lewis structures for thecarbonate ion and for phosphine in which the central atom obeys the octet rule. ... How many equivalent Lewis structures are necessary to describe the bonding in CO32-
Answer:
See explanation
Explanation:
A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.
Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.
The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.
A student accidentally let some of the vapor escape the beaker. As a result of this error, will the mass of naphthalene you record be too high, too low, or unaffected? Why?
Answer:
too low
Explanation:
If our aim is to recover the naphthalene and measure its mass after separation, then we must not allow any vapour to escape.
Naphthalene is a sublime substance, it can be separated by sublimation. It changes directly from solid to gas. This vapour must be kept securely so that none of it escapes. If part of the naphthalene vapour happens to escape accidentally, then the measured mass of naphthalene will be too low compared to the mass of naphthalene originally present in the mixture.