Answer: [tex]2AgNO_3(aq)+BaCl_2(aq)\rightarrow 2AgCl(s)+Ba(NO_3)_2(aq)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
The reaction for aqueous silver nitrate solution with barium chloride solution produces barium nitrate solution and a white precipitate of silver chloride.
Thus the balanced chemical reaction will be:
[tex]2AgNO_3(aq)+BaCl_2(aq)\rightarrow 2AgCl(s)+Ba(NO_3)_2(aq)[/tex]
the molar solubility of Zn(OH)2 is 5.7x 10^-3 mol/L at a certain temperature. Calculate the value of Ksp for Zn(OH)2 at this temperataure
Answer:
Ksp = 7.4x10⁻⁷
Explanation:
Molar solubility of a substance is defined as the amount of moles of that can be dissolved per liter of solution.
Ksp of Zn(OH)₂ is:
Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻
Ksp = [Zn²⁺] [OH⁻]²
And the molar solubility, X, is:
Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻
⇄ X + 2X
Because X are moles of substance dissolved.
Ksp = [X] [2X]²
Ksp = 4X³
As molar solubility, X, is 5.7x10⁻³mol/L:
Ksp = 4X³
Ksp = 4 (5.7x10⁻³mol/L)³
Ksp = 7.4x10⁻⁷When H2O and CO react at 979°C, the products are CO2 and H2. The equilibrium constant (in terms of equilibrium concentrations of reactants and products) for the reaction below is 0.66 at 979°C. If the following concentrations are measured after the reaction reaches equilibrium, what is the concentration of CO(g) in the equilibrated mixture? answer will be in M
Component: Measured Equilibrium Concentration
A. H2 0 (g) 0.750 M
B. CO2 (g) 0.134 M
C. H2 (g) 3.33 M
Answer:
0.901 M
Explanation:
The concentration of the CO(g) in the equilibrated mixture is shown below:
[tex]K_C = \frac{H_2\times CO_2}{CO\times H_2O}[/tex]
where,
[tex]K_C[/tex] = equilibrium constant
And, we placing these above values to the formula shown as above
So, the concentration of CO(g) is
[tex]= \frac{3.33 \times 0.134}{0.66\times 0.750}[/tex]
= [tex]\frac{0.44622}{0.495}[/tex]
= 0.901 M
We simply applied the above formula in order to determine it and so that the correct answer could arrive
Which Group has 1 valence electron?
A. Alkali metals
B. Lanthanides
C. Transition metals
D. Alkaline earth metals
Answer:
A
Explanation:
The group that has one valence electron is the first group, also known as the alkali metals.
The Ksp of calcium sulfate, CaSO4, is 9.0 × 10-6. What is the concentration of CaSO4 in a saturated solution? A. 3.0 × 10-3 Molar B. 9.0 × 10-3 Molar C. 3.0 × 10-6 Molar D. 9.0 × 10-6 Molar
Answer: The concentration of [tex]CaSO_4[/tex] in a saturated solution is [tex]3.0\times 10^{-3}M[/tex]
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
The equation for the ionization of [tex]CaSO_4[/tex] is given as:
[tex]K_{sp}[/tex] of [tex]CaSO_4[/tex] = [tex]9.0\times 10^{-6}[/tex]
By stoichiometry of the reaction:
1 mole of [tex]CaSO_4[/tex] gives 1 mole of [tex]Ca^{2+}[/tex] and 1 mole of [tex]SO_4^{2-}[/tex]
When the solubility of [tex]CaSO_4[/tex] is S moles/liter, then the solubility of [tex]Ca^{2+}[/tex] will be S moles\liter and solubility of [tex]SO_4^{2-}[/tex] will be S moles/liter.
[tex]K_{sp}=[Ca^{2+}][SO_4^{2-}][/tex]
[tex]9.0\times 10^{-6}=[s][s][/tex]
[tex]9.0\times 10^{-6}=s^2[/tex]
[tex]s=3.0\times 10^{-3}M[/tex]
Thus concentration of [tex]CaSO_4[/tex] in a saturated solution is [tex]3.0\times 10^{-3}M[/tex]
1. List the conjugate acid or conjugate base for each chemical. a. The acid HF b. The base KOH c. The base NH3 d. The acid HNO3 e. The acid HCOOH f. The base CH3NH2
Answer:
a) Conjugate base F– b) Conjugate acid K+ c) Conjugate acid NH4+ d) Conjugate base NO2- e) Conjugate base HCOO– f) Conjugate acid CH4+
Explanation:
Acid will produce Conjugate base
Base will produce Conjugate acid.
Answer:
a. The acid HF: F-
b. The base KOH: H2O
c. The base NH3: NH4+
d. The acid HNO3: NO3-
e. The acid HCOOH: COOH-
f. The base CH3NH2: CH3NH3+
Explanation:
A sample of an unknown substance has a mass of 0.158kg. If 2,520.0 j of heat is required to heat the substance from 32.0C to 61.0C what is the specific heat of the substance
Calculate the number of milliliters of 0.656 M KOH required to precipitate all of the Co2 ions in 187 mL of 0.745 M Co(NO3)2 solution as Co(OH)2. The equation for the reaction is:
Answer:
THE MILLILITERS OF 0.656 M KOH REQUIRED TO PRECIPITATE ALL THE Co2 IONS IN 187 mL OF 0.745 M Co(NO3)2 SOLUTION IS 212.37 mL
EQUATION FOR THE REACTION IS :
2 KOH + Co(NO3)2 ----------> Co(OH)2 + 2 KNO3
Explanation:
Using dilution formula:
M1V1 = M2V2
V2 = M1 V1 / M2
M1 = 0.745 M
V1 = 187 mL
M2 = 0.656 M
V2 = unknown
V2 = 0.745 * 187 / 0.656
V2 = 139.315 / 0.656
V2 = 212.37 mL
the number of milliliters of 0.656 M KOH required to precipitate all of the Co 2 ions is 212.37 mL.
The equation for the reaction is:
2KOH + Co(NO3)2 ----------> Co(OH)2 + 2KNO3
That is 2 moles of potassium hydroxide react with 1 mole of cobalt(11) nitrate to form 1 mole of cobalt hydroxide and 2 moles of potassium nitrate
How many moles of sulfur trioxide will be produced when the complete combustion of 100.0 g of sulfur dioxide takes place
Answer:
1.563 moles of SO3.
Explanation:
We begin by calculating the number of mole present in 100g of sulphur dioxide, SO2. This can be obtained as follow:
Molar mass of SO2 = 32 + (16x2) = 64g/mol
Mass of SO2 = 100g
Mole of SO2 =..?
Mole = mass/Molar mass
Mole of SO2 = 100/64
Mole of SO2 = 1.563 mole
Now, we can obtain the number of mole of sulphur trioxide, SO3 produce from the reaction as follow:
2SO2 + O2 —> 2SO3
From the balanced equation above,
2 moles of SO2 reacted to produce 2 moles of SO3.
Therefore, 1.563 moles of SO2 will also react to produce 1.563 moles of SO3.
Therefore, 1.563 moles of SO3 is obtained from the reaction.
If you needed a 1.5 x 1 0-4 M solution of a compound that has a molar mass of 760 g/mol, what would it concentration be in parts per million?
Answer:
114 ppm
Explanation:
Data obtained from the question include:
Conc. of compound in mol/L = 1.5×10¯⁴ mol/L
Molar mass of compound = 760 g/mol
Conc. in ppm =..?
Next, we shall determine the concentration of the compound in grams per litre (g/L) . This is illustrated below:
Conc. in mol/L = conc. in g/L / Molar mass
1.5×10¯⁴ = conc. In g/L / 760
Cross multiply
Conc. in g/L = 1.5×10¯⁴ x 760
Conc. in g/L = 0.114 g/L
Next, we shall convert 0.114 g/L to milligrams per litre (mg/L). This is illustrated below:
1 g/L = 1000 mg/L
Therefore, 0.114 g/L = 0.114 x 1000 = 114 mg/L
Finally, we shall convert 114 mg/L to parts per million (ppm). This is illustrated below:
1 mg/L = 1 ppm
Therefore, 114 mg/L = 114 ppm
From the calculations made above,
1.5×10¯⁴ mol/L Is equivalent to 114 ppm.
Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16.0 mL of CO?
2 CO(g) + O2(g)
2 CO2(g)
Express your answer with the appropriate units.
Answer:
[tex]V_{CO_2}=16.0mL[/tex]
Explanation:
Hello,
In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:
[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]
We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:
[tex]n_{CO}V_{CO}=n_{CO_2}V_{CO_2}[/tex]
Thus, solving for the yielded volume of carbon dioxide we obtain:
[tex]V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL[/tex]
Best regards.
The average human body contains 5.00 L of blood with a Fe2+ concentration of 1.10×10−5 M . If a person ingests 9.00 mL of 21.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?
Answer:
The percentage is % [tex]Fe^{2+[/tex] [tex]= 57.3[/tex]%
Explanation:
From the question we are told that
The volume of blood in the human body is [tex]V = 5.0 0 \ L[/tex]
The concentration of [tex]Fe^{2+[/tex] is [tex]C_{F} = 1.10 *10^{-5} \ M[/tex]
The volume of NaCN ingested is [tex]V_N = 9.00 \ mL = 9.00 *10^{-3} \ L[/tex]
The concentration of NaCN ingested is [tex]C_N = 21.0 \ mM = 21.0 *10^{-3} \ M[/tex]
The number of moles of [tex]Fe^{2+[/tex] in the blood is
[tex]N_F = C_F * V[/tex]
substituting values
[tex]N_F = 1.10 *10^{-5} * 5[/tex]
[tex]N_F = 5.5*10^{-5} \ mols[/tex]
The number of moles of [tex]CN^{-}[/tex] ingested is mathematically evaluated as
[tex]N_C = C_N * V_N[/tex]
substituting values
[tex]N_C = 21*10^{-3} * 9 *10^{-3}[/tex]
[tex]N_C = 1.89 *10^{-4} \ mols[/tex]
The balanced chemical equation for the reaction between [tex]Fe^{2+[/tex] and [tex]CN^{-}[/tex] is represented as
[tex]Fe^{2+} + 6 CN^{-} \to [Fe(CN)_6]^{2-}[/tex]
From this reaction we see that
1 mole of [tex]Fe^{2+[/tex] will react with 6 moles of [tex]CN^{-}[/tex]
=> x moles of [tex]Fe^{2+[/tex] will react with [tex]1.89 *10^{-4} \ moles[/tex] of [tex]CN^{-}[/tex]
Thus
[tex]x = \frac{1.89 *10^{-4} * 1}{6}[/tex]
[tex]x = 3.15 *10^{-5}[/tex]
Hence the percentage of [tex]Fe^{2+[/tex] that reacted is mathematically evaluated as
% [tex]Fe^{2+[/tex] [tex]= \frac{3.15 *10^{-5}}{5.5*10^{-5}} * 100[/tex]
% [tex]Fe^{2+[/tex] [tex]= 57.3[/tex]%
Given the information about each pair of acids fill in the correct answer.
a. Acid A has a lower % ionization than B:_______ is a stronger acid.
b. Acid B has a larger K_a than acid A._______ will have a larger percent ionization.
c. A is a stronger acid than B. Acid B will have________ percent ionization than A.
Answer:
a. Acid B
b. Acid B
c. lower
Hope this helps you
How much energy in joules will be required to raise the temperature of 50.0 g of water from 20 degrees C to 60 degree C
Answer: 8368 Joules
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed or released =?
c = specific heat capacity of water = [tex]4.184J/g^0C[/tex]
Initial temperature of water = [tex]T_i[/tex] = [tex]20^0C[/tex]
Final temperature of water = [tex]T_f[/tex] = [tex]60^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(60-20)^0C=40^0C[/tex]
Putting in the values, we get:
[tex]Q=50.0g\times 4.184J/g^0C\times 40^0C=8368J[/tex]
Thus energy in Joules required is 8368.
Which of the following cannot have hydrogen bonds? Select one: A. NH3 B. H2O C. HF D. CH3NH2 E. Which of the following cannot have hydrogen bonds? Select one: A. NH3 B. H2O C. HF D. CH3NH2 E. HCl
Answer:
E. HCl
Explanation:
Cl atom does not have enough electronegativity to make enough positive charge on H.
HCl is the compound which doesn't have hydrogen bonds. This is because of
the higher size of the chlorine atom.
There is no hydrogen bond because of the high size of the chlorine.
Chlorine have electrons with a very low density. It is also very
electronegative which explains why the formation of hydrogen bonds in the
compound HCl is not possible.
Instead, HCl has covalent bonds in which electron is shared between the
hydrogen and chlorine to achieve a stable configuration.
Read more on https://brainly.com/question/10462894
Write empirical formula
Answer:
[tex]Pb(ClO_{3})_{4}\\Pb(MnO_{4})_{4}\\Fe(ClO_{3})_{3}\\\Fe(MnO_{4})_{3}\\[/tex]
Explanation:
[tex]Pb^{4+}(ClO_{3}^{-})_{4}--->Pb(ClO_{3})_{4}\\Pb^{4+}(MnO_{4}^{-})_{4}--->Pb(MnO_{4})_{4}\\Fe^{3+}(ClO_{3}^{-})_{3}--->Fe(ClO_{3})_{3}\\\Fe^{3+}(MnO_{4}^{-})_{3}--->Fe(MnO_{4})_{3}\\[/tex]
8. How many grams of SO2 are there in 2.5 mol of SO2? (Show Work)
Answer:
160g
Explanation:
Mass in grams is equal to product of moles and molar mass of compound.
Please help me solve this it’s very important I get this right
Answer:
D. exothermic reaction
Explanation:
In an exothermic reaction, the reactants are at a higher energy level than the products.
How are the oxygen atoms balanced for redox equations in basic solutions
Answer: H2O and OH^- are added to balance the oxygen atoms.
Explanation:
The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. How much tritium will be left after 30 years
Explanation:
Half life = 12.3years
Time = 30 years
Basically half life is the amount of time taken for the intial concentration to be reduced to half.
First half life = 12.3 years = 10/2 = 5 Kg left
Second half life = 24.6 years= 5/2 = 2.5 Kg left
Third half life = 36.9 years = 2.5 / 2 = 1.25 Kg left
This means that after 30 years, the amount of tritium left would e betweem 1.25kg to 2.5 kg.
The number of tritium that will be left after 30 years is 1.844.
Calculation of the number of tritium left:The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance Also the number of years should be 30 years
So,
= 10*2^(-30/12.3)
= 1.844
Therefore, we can conclude that The number of tritium that will be left after 30 years is 1.844.
Learn more about substance here: https://brainly.com/question/24840674
Calculate the pH and concentrations of H2A, HA−, and A2−, at equilibrium for a 0.236 M solution of Na2A. The acid dissociation constants for H2A are Ka1=7.68×10−5 and Ka2=6.19×10−9.
Answer:
[H₂A] = 5.0409x10⁻⁷M
[HA⁻] = 0.001951M
[A²⁻] = 0.234
11.29 = pH
Explanation:
When Na₂A is in equilibrium with water, the reactions that occurs are:
2Na⁺ + A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + 2Na⁺(aq) + OH⁻(aq)
As sodium ion doesn't react:
A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)
Kb1 = KwₓKa2 = 1x10⁻¹⁴/ 6.19x10⁻⁹ = 1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]
And HA⁻ will be in equilibrium:
HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)
Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ = 1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]
In the reaction, you have 2 equilibriums, for the first reaction, concentrations in equilibrium are:
[HA⁻] = X
[OH⁻] = X
[A²⁻] = 0.236M - X
Replacing in Kb1:
1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]
1.6155x10⁻⁶ = [X] [X] / [0.236-X]
3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²
3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0
Solving for X
X = -0.00195 → False solution. There is no negative concentrations
X = 0.001952.
Replacing, concentrations for the first equilibrium are:
[HA⁻] = 0.001952
[OH⁻] = 0.001952
[A²⁻] = 0.234
Now, in the second equilibrium:
[HA⁻] = 0.001952 - X
[OH⁻] = X
[H₂A] = X
Replacing in Kb1:
1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]
1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]
2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²
2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0
Solving for X
X = -5.04x10⁻⁷ → False solution. There is no negative concentrations
X = 5.0409x10⁻⁷
Replacing, concentrations for the second equilibrium are:
[HA⁻] = 0.001951M
[OH⁻] = 5.0409x10⁻⁷M
[H₂A] = 5.0409x10⁻⁷M
Thus, you have concentrations of H2A, HA−, and A2−
Now, for pH, the sum of both productions of [OH⁻] is:
[OH⁻] = 0.0019525
pOH = -log[OH⁻] = 2.709
As 14 = pH+ pOH
11.29 = pH
As per the question the pH and the cons of the H2A, the HA−, and the A2−, at equilibrium for a 0.236 M.
The pH needs to be in equilibrium from the mentioned elements and form a solution of Na2A. Thus the concentration of the ions is to be calculated with the dissociation of the constants for the H2A.Hence the [H₂A] = 5.0409x10⁻⁷M.[HA⁻] = 0.001951M A²⁻] = 0.234 will give 11.29 = pH.Learn more about the A2−, at equilibrium.
brainly.com/question/17086012.
Which are processes that add to the genetic differences in siblings? Cheek all that apply. Interphase Independent assortment Cytokinesis. Crossing over Mitosis
Answer:crossing over
Explanation:
Answer:
Crossing Over and Independent Assortment.
Explanation:
Crossing over: In Prophase I of Meiosis I, homologous chromosomes line up their chromatids and "cross-over", or exchange corresponding segments of DNA with each other. This produces genetic variation by allowing more combinations of genes to be produced.
Independent Assortment: In Anaphase I of Meiosis I, homologues separate and move to opposite sides of the cell. Resulting cells have one chromosome from each pair of homologous chromosomes. However, WHICH chromosome that each cell gets is completely random.
A principal constituent of petrol (gasoline) is iso-octane, C8H18. From the following thermodynamic data at
298 K what is the
standard molar enthalpy of combustion of iso-octane in excess oxygen
at 298 K?
C«H;8(1) + 12702() +8C02(g) +91,0(1)
Substance AfHn/kJ mol"}
C8H8(1)
-258.07
02(8)
0
CO2(8)
-393.51
H2O(1)
-285.83
Answer: The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol
Explanation:
The balanced reaction for combustion of isooctane is:
[tex]C_8H_{18}(l)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{(CO_2(g))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_{C_8H_{18}(l)}=-258.07kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(8\times (-393.51))+(9\times (-285.8))]-[(1\times (-258.07))+(\frac{25}{2}\times (0))]\\\Delta H^o_{rxn}=-5462.2kJ/mol[/tex]
The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol
Which of the following is a likely mechanism for the reaction CH3Cl + OH- ----> CH3OH + Cl-, which is first order with respect to each of the reactants? A one-step mechanism involving a transition state that contains two hydroxide ions "attached" to the carbon atom of CH3Cl A one-step mechanism involving a transition state that has a carbon partially bonded to both chlorine and oxygen A two-step mechanism in which the chlorine leaves CH3Cl in a slow step, followed by rapid attack of the intermediate by the hydroxide ion A two-step mechanism in which the chlorine leaves CH3Cl in a rapid step, followed by the slow attack of the intermediate by the hydroxide ion
Answer:
A one-step mechanism involving a transition state that has a carbon partially bonded to both chlorine and oxygen
Explanation:
The compound CH3Cl is methyl chloride. This is a nucleophilic substitution reaction that proceeds by an SN2 mechanism. The SN2 mechanism is a concerted reaction mechanism. This means that the departure of the leaving group is assisted by the incoming nucleophile. The both species are partially bonded to opposite sides of the carbon atom in the transition state.
Recall that an SN2 reaction is driven by the attraction between the negative charge of the nucleophile (OH^-) and the positive charge of the electrophile (the partial positive charge on the carbon atom bearing the chlorine leaving group).
Combustion of 30.42 g of a compound containing only carbon, hydrogen, and oxygen produces 35.21 g CO2 and 14.42 g H2O. What is the empirical formula of the compound
Answer:
C2H4O3
Explanation:
We would have to do some preparations between before solving it the normal way. The main goal is to get the masses of the Individual elements. So here goes;
We can get the mass of C from CO2 using the following steps:
1 mole of CO2 has a mass of 44g (Molar mass) and contains 12g of C.
How did we know the molar mass of CO2 is 44g?
Easy. 1 mole of C = 12, 1 mole of O = 16
But we have two O’s so the total mass of O = (2 * 6) = 32
Total mass of CO2 = mass of C + Mass of O = 12 + 32 = 44
So if 44g of CO2 contains 12g of C, how much of C would be present in 35.21g CO2.
12 = 44
X = 35.21
X = (35.21 * 12) / 44 = 9.603g
We can also get the mass of H from H2O. 1 mole of H2O has a mass of 18g and contains 2g of H.
How did we know the molar mass of H2O is 18g?
Easy. 1 mole of H = 1, 1 mole of O = 16
But we have two H’s so the total mass of H = (2 * 1) = 2
Total mass of H2O = mass of H + Mass of O = 2 + 16 = 18
So how much of H would be present in 14.42g of H2O?
2 = 18
X =14.42
X = (14.42 * 2 ) / 18 = 1.602g
Now we have the masses of C and H. But the question says the compound contains the C, H and O.
So we still have to calculate the mass of Oxygen. We obtain this from;
Mass of Compound = Mass of Carbon + Mass of Oxygen + Mass of Hydrogen
Mass of Oxygen = Mass of compound – (Mass of Carbon + Mass of Hydrogen)
Mass of Oxygen = 30.42 – (9.603 + 1.602)
Mass of Oxygen = 30.42 - 11.205 = 19.215
Now we have all the masses so we are good too go. Let’s have our table.
Elements Carbon (C) Hydrogen (H) Oxygen (O)
Mass 9.603 1.602 19.215
0.800 1.602 1.2001 (Divide by molar mass)
1 2 1.5 (Divide by lowest number)
2 4 3 (Convert to simple integers by * 2)
The Empirical formula of the compound is C2H4O3
An aqueous solution was made by dissolving 72.9 grams of glucose, C6H12O6, into 115 grams of water. The vapor pressure of the pure water is 26.4 Torr. The vapor pressure of water over this solution is: (a) 27.9 Torr (b) 24.1 Torr (c) 26.8 Torr (d) 24.8 Torr PLease answer this as quick as possible
Answer:
The correct answer is (d) 24.8 Torr
Explanation:
When a solute is added to a solvent, the water pressure of the solution is lower than the vapor pressure of the pure solvent. This is called vapor pressure lowering and it is given by the following expression:
Psolution= Xsolvent x Pºsolvent
We have to calculate Xsolvent (mole fraction of solvent) which is given by the number of moles of solute divided into the total number of moles.
First, we calculate the number of moles of solute and solvent. The solute is glucose (C₆H₁₂O₆), and its number of moles is calculated from the mass and the molecular weight (MM):
MM (C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol
moles of glucose= mass/MM= (72.9 g)/)(180 g/mol)= 0.405 moles
The solvent is water (H₂O) and again we calculate the number of moles as follows:
MM(H₂O)= (1 g/mol x 2) + 16 g/mol = 18 g/mol
moles of water= mass/MM= (115 g)/(18 g/mol)= 6.389 moles
Now, we calculate the total number of moles (nt):
nt= moles of glucose + moles of water= 0.405 moles + 6.389 moles= 6.794 moles
The mole fraction of water (Xsolvent) is given by:
Xsolvent= moles of water/nt= 6.389 moles/6.794 moles= 0.940
Finally, the vapor pressure of water over the solution will be the following:
Psolvent= Xsolvent x Pºsolvent= 0.940 x 26.4 Torr= 24.8 Torr
Potassium, a metal with one electron in the outermost shell, will react with how many chlorine atoms
Answer:
7 chlorine atoms
Explanation:
K=2.8.8.1
Cl=2.8.7
pottasium will give chlorine its I valence electron to form ions as follows
K=(2.8.8)+
Cl=(2.8.8)-
It will react with 1 chlorine atom.
Whilst one atom loses an electron to every other atom it results in the formation of?
An ionic bond is shaped by using the whole transfer of some electrons from one atom to every other. The atom losing one or more electrons becomes a cation—an undoubtedly charged ion. The atom gaining one or more electrons will become an anion—a negatively charged ion.
What number of bonds can chlorine form?In those compounds carbon, nitrogen, oxygen, and chlorine atoms have four, three, and one bonds, respectively. The hydrogen atom and the halogen atoms form the most effective covalent bond to different atoms in maximum stable neutral compounds.
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The complete combustion of ethanol, C2H5OH(l), to form H2O(g) and CO2(g) at constant pressure releases 1235 kJ of heat per mole of C2H5OH.
Write a balanced equation for this reaction.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Calculate the boiling point of a solution prepared by adding 11.5 g naphthalene (C10H8) to 250.0 g of benzene. Naphthalene is a non-electrolyte solute, and benzene is an organic solvent that exhibits a boiling point of 80.10 oC, and has a kb
Answer:
81.0°C
Explanation:
Kb benzene = 2.5°C/m
The addition of a solute to a pure solvent produce an elevation in boiling point regarding to boiling point of pure solvent. This phenomenon follows the equation:
ΔT = Kb×m×i
where ΔT represents the increasing in boiling point, Kb is the elevation boiling point constant of the solvent (2.5°C/m for benzene), m is molality of solution (Moles solute / kg solvent) and i is Van't Hoff factor (1 for a non-electrolyte solute as naphthalene).
Moles of 11.5g of naphthalene (Molar mass: 128.17g/mol) are:
11.5g × (1mol / 128.17g) = 0.0897 moles of naphthalene in 250.0g = 0.250kg of benzene.
Molality is:
0.0897 moles of naphthalene / 0.250kg of benzene = 0.359m
Replacing in the equation:
ΔT = Kb×m×i
ΔT = 2.5°C/m×0.359m×1
ΔT = 0.90°C
That means the solution prepared has an elevation in boiling point of 0.90°C. As boiling point of pure benzene is 80.10°C, boiling point of the solution is:
80.10°C + 0.90°C =
81.0°CA chemist has a block of aluminum metal (density is 2.7 g/mL). The block weighs 1.5. What is the volume of the aluminum block?
Answer:
0.56 mL
Explanation:
Volume = mass ÷ density
Volume = 1.5 ÷ 2.7 g/mL
Volume = 0.5555555556 = 0.56 mL
The volume of the aluminum block is 0.56 mL.
Hope this helps. :)
The volume of aluminum block is 0.556 mL.
The density of a substance is its mass per unit volume.
It given by formula:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} [/tex]
Given:
Density = 2.7 g/mL
Mass= 1.5 g
To find:
Volume=?
On substituting the values in the above formula:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} \\\\\text{Volume}=\frac{\text{Mass}}{\text{Density}} \\\\\text{Volume}=\frac{1.5}{2.7} \\\\\text{Volume}=0.556mL[/tex]
Thus, the volume of the aluminum block is 0.556mL.
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Nitrogen monoxide reacts with chlorine at high temperature according to the equation, 2 NO(g) + Cl2(g) → 2 NOCl(g) In a certain reaction mixture the rate of formation of NOCl(g) was found to be 4.50 x 10‑4 mol L‑1 s‑1. What is the rate of consumption of NO(g)?
Answer:
4.50 × 10⁻⁴ mol L⁻¹ s⁻¹
Explanation:
Step 1: Write the balanced equation
2 NO(g) + Cl₂(g) → 2 NOCl(g)
Step 2: Establish the appropriate molar ratio
The molar ratio of NO(g) to NOCl(g) is 2:2, that is, when 2 moles of NO(g) are consumed, 2 moles of NOCl(g) are formed.
Step 3: Calculate the rate of consumption of NO(g)
The rate of formation of NOCl(g) is 4.50 × 10⁻⁴ mol L⁻¹ s⁻¹. The rate of consumption of NO(g) is:
[tex]\frac{4.50 \times 10^{-4}molNOCl}{L.s} \times \frac{2molNO}{2molNOCl} = \frac{4.50 \times 10^{-4}molNO}{L.s}[/tex]