First you need to calculate the number of moles of aluminium and copper chloride.
number of moles = mass / molecular weight
moles of Al = 512 / 27 = 19 moles
moles of CuCl = 1147 / 99 = 11.6 moles
From the reaction you see that:
if 2 moles of Al will react with 3 moles of CuCl
then 19 moles of Al will react with X moles of CuCl
X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.
Which of the following results from acid deposition? View Available Hint(s) Which of the following results from acid deposition? Chlorofluorocarbons are formed near the ground. Species diversity is increased. Objects made of stone, like marble headstones, are worn away. Food supply for birds is increased.
Answer:
Objects made of stone, like marble headstones, are worn away.
Explanation:
Acid deposition on earth is mostly as a result of acid rain. Acid rain forms which oxides of non-metals such as nitrogen and carbon dissolves in rain water to produce weak acid.
This weak acid that is produced easily wears away materials made up of marble and other sculptures. The most active of the acid is weak carbonic acid produced by dissolution of carbon dioxide in rain water.
Acid deposition leads to the deterioration of objects made of stone, such as marble headstones, because of the corrosion of metals and the wearing away of stone due to acidic precipitation.
Acid deposition results from the chemical reaction of air pollutants, like sulfur dioxide (SO2) and nitrogen oxides (NOx), with the atmosphere to form acids such as nitric and sulfuric acids. These acidic compounds can fall to Earth in various forms, either as wet precipitation like rain, sleet, or snow, or as dry particles. The impacts of acid deposition are numerous and widespread, including ecological and material damage.
One of the notable consequences of acid deposition is the deterioration of objects made of stone, such as marble headstones. Acid rain and acid particles contribute to the corrosion of metals and the deterioration of paint and stone, including culturally significant statues, monuments, and other stonework. This corrosive effect leads to the loss of structural integrity and the diminishment of aesthetic and historical value.
Therefore, among the listed options, the result of acid deposition is that objects made of stone, like marble headstones, are worn away.
Identify the Brønsted-Lowry acid, the Brønsted-Lowry base, the conjugate acid, and the conjugate base in each reaction:
(a) C5H5N(aq)+H2O(l)→C5H5NH+(aq)+OH−(aq)
(b) HNO3(aq)+H2O(l)→H3O+(aq)+NO3−(aq)
Answer:
C5H5N is the base and C5H5NH+ is the conjugate acid
H2O is the acid and OH− is the conjugate base
Explanation:
Hydrogen + is also called a proton
C5H5N is the base because it receives the proton (H+) and C5H5NH+ is its conjugate acid
H2O is the acid because it gives up the proton and OH− is the conjugate base because it is capable of receiving the proton
Answer:
HNO3 is the acid and NO3- is the conjugate base
H2O is the base and H3O+ is the conjugate acid
Explanation
HNO3 is the acid and NO3− is its conjugate base, capable of receiving a proton
H2O is the base because it receives the proton and H3O+ is a conjugate acid capable of giving up the proton.
In the reactions given, C5H5N acts as the Brønsted-Lowry base and H2O as the acid in the first reaction, forming C5H5NH+ as the conjugate acid and OH- as the conjugate base. In the second reaction, HNO3 is the acid and H2O is the base, yielding H3O+ as the conjugate acid and NO3- as the conjugate base.
Explanation:According to the Brønsted-Lowry theory, an acid is a substance that donates a proton (H+), and a base is a substance that accepts a proton. When an acid donates a proton, it forms its conjugate base, and when a base accepts a proton, it forms its conjugate acid. In the provided reactions:
(a) C5H5N(aq) is the Brønsted-Lowry base (BB) as it accepts a proton, H2O(l) is the Brønsted-Lowry acid (BA) as it donates a proton, C5H5NH+(aq) is the conjugate acid (CA) of the pyridine after accepting a proton, and OH−(aq) is the conjugate base (CB) of water after donating a proton.(b) HNO3(aq) is the Brønsted-Lowry acid (BA) as it donates a proton, H2O(l) is the Brønsted-Lowry base (BB) as it accepts a proton, H3O+(aq) is the conjugate acid (CA) of the water after accepting a proton, and NO3−(aq) is the conjugate base (CB) of the nitric acid after donating a proton.Which process most likely takes place when the kinetic energy of molecules overcomes the intermolecular forces?
This is melting or boiling
Solute crystals are added to a solution until they begin to sink to the bottom of their container. The solution is then heated, and many more solute crystals dissolve. Which term best describes the solution immediately after it was heated?A. saturatedB. unsaturatedC. concentratedD. supersaturated
Answer:
D. supersaturated
Explanation:
The solution is said to be supersaturated. Generally, the rate of solubility of substances can be increased by grinding, stirring and heating.
For a solution that contains enough solute than it can normally dissolve at a particular temperature, we say it is supersaturated.
The point where the additional solute crystals begins to sink to the bottom of the container points to the fact that the solution is saturated already and would no more dissolve any more solute. When the solution becomes heated, dissolution is induced and more solute begins to dissolve, this makes the solution supersaturated. This solution now contains more solute than it can normally hold.
Answer:
B. unsaturated
Explanation:
Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO and O2 are separated in two different chambers connected by a valve. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temperature remains constant at 25°C. Initial conditions are as follows: NO: 3.90 L, 0.500 atm O2: 2.09 L, 1.00 atm
Answer:
The remain gases are [tex]o_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]
Pressure of [tex]O_{2}_{(g)}[/tex] [tex]1.09 atm O_{2}_{(g)}[/tex]
Pressure of [tex]NO_{2}_{(g)}[/tex] [tex]1.09 atm NO_{2}_{(g)}[/tex]
Explanation:
We have the following reaction
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]
Now we calculate the limit reagents, to know which of the two gases is completely depleted and which one is in excess.
Excess gas will remain in the tank when the reagent limits have run out and the reaction ends.
To calculate the limit reagent, we must calculate the mols of each substance. We use the ideal gas equation
[tex]PV= nRT[/tex]
We cleared the mols
[tex]n=\frac{PV}{RT}[/tex]
PV=nrT
replace the data for each gas
Constant of ideal gases
[tex]R= 0.082\frac{atm.l}{mol.K}[/tex]
Transform degrees celsius to kelvin
[tex]25+273=298K[/tex]
[tex]NO_{g}[/tex]
[tex]n=\frac{0.500atm.3.90l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.080mol NO_{(g)} \\[/tex]
[tex]O_{2}_{(g)[/tex]
[tex]n=\frac{1atm.2.09l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.086mol O_{2}_{(g)} \\[/tex]
Find the limit reagent by stoichiometry
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]
[tex]0.086mol O_{2}_(g).\frac{2mol NO_{(g)} }{1mol O_{2}_{(g)} } =0.17mol NO_{(g)}[/tex]
Using [tex]O_{2}_{(g)}[/tex]as the limit reagent produces more [tex]NO_{(g)}[/tex] than I have, so oxygen is my excess reagent and will remain when the reaction is over.
[tex]NO_{(g)}[/tex]
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{1mol O_{2}_ {(g)} }{2mol NO_{(g)} } =0.04mol O_{2}_{(g)}[/tex]
Using [tex]NO_{(g)}[/tex] as the limit reagent produces less [tex]O_{2}_{(g)}[/tex] than I have, so [tex]NO_{(g)}[/tex] is my excess reagent and will remain when the reaction is over.
Calculate the moles that are formed of [tex]NO_{2}_{(g)}[/tex]
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{2mol NO_{2}_ {(g)} }{2mol NO_{(g)} } =0.080mol NO_{2}_{(g)}[/tex]
We know that for all [tex]NO_{(g)}[/tex] to react, 0.04 mol [tex]O_{2}_{(g)}[/tex] is consumed.
we subtract the initial amount of [tex]O_{2}_{(g)}[/tex] less than necessary to complete the reaction. And that gives us the amount of mols that do not react.
[tex]0.086-0.04= 0.046[/tex]
The remain gases are[tex]O_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]
calculate the volume that gases occupy
[tex]0.080 mol NO_{2}_{(g)} .\frac{22.4l NO_{2}_{(g)} }{1molNO_{2}_{(g)} }} =1.79 lNO_{2}_{(g)}[/tex]
[tex]0.046 mol O_{2}_{(g)} .\frac{22.4 l O_{2}_{(g)} }{1molO_{2}_{(g)} }} =1.03 l O_{2}_{(g)}[/tex]
Calculate partial pressures with the ideal gas equation
[tex]PV= nRT[/tex]
[tex]P=\frac{nRT}{V}[/tex]
Pressure of [tex]O_{2}_{(g)}[/tex]
[tex]P=\frac{0.046mol.0.082\frac{atm.l}{K.mol} 298K}{1.03l}= 1.09 atmO_{2}_{(g)}[/tex]
Pressure of [tex]NO_{2}_{(g)}[/tex]
[tex]P=\frac{0.080mol.0.082\frac{atm.l}{K.mol} 298K}{1.79l}= 1.09 atmNO_{2}_{(g)}[/tex]
Answer:
It will remain O₂ and NO₂, with partial pressures: pO₂ = 0.186 atm, and pNO₂ = 0.325 atm.
Explanation:
First, let's identify the initial amount of each reactant using the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas equation (0.082 atm.L/mol.K), and T is the temperature (25°C + 273 = 298 K).
n = PV/RT
NO:
n = (0.500*3.90)/(0.082*298)
n = 0.0798 mol
O₂:
n = (1.00*2.09)/(0.082*298)
n = 0.0855 mol
By the stoichiometry of the reaction, we must found which reactant is limiting and which is in excess. The limiting reactant will be totally consumed. Thus, let's suppose that NO is the limiting reactant:
2 moles of NO ------------------ 1 mol of O₂
0.0798 mol ------------------- x
By a simple direct three rule:
2x = 0.0798
x = 0.0399 mol of O₂
The number of moles of oxygen needed is lower than the number of moles in the reaction, so O₂ is the limiting reactant, and NO will be totally consumed. The number of moles of NO₂ formed will be:
2 moles of NO --------------- 2 moles of NO₂
0.0798 mol ---------------- x
By a simple direct three rule:
x = 0.0798 mol of NO₂
And the number of moles of O₂ that remains is the initial less the total that reacts:
n = 0.0855 - 0.0399
n = 0.0456 mol of O₂
The final volume will be the total volume of the containers, V = 3.90 + 2.09 = 5.99 L, so by the ideal gas law:
PV = nRT
P = nRT/V
O₂:
P = (0.0456*0.082*298)/5.99
P = 0.186 atm
NO₂:
P = (0.0798*0.082*298)/5.99
P = 0.325 atm
PLEASE ANSWER QUICKLY!
How many grams of H3PO4 are produced when 12.81 moles of water react with an excess of P4O10?
Select one:
a. 837
b. 1880
c. 471
d. 8.54
Answer:
c
c
Explanation:
[tex]h3po4 + 12.81 - p40.10[/tex]
Answer:
A. 837 grams of [tex]H_3PO_4[/tex] are produced when 12.81 moles of water react with an excess of [tex]P_4O_{10}[/tex]
Explanation:
The balanced chemical equation is
[tex]P_4 O_10 + 6 H_2 O > 4 H_3 PO_4[/tex]
Mole ratio of [tex]H_2 O:H_3 PO_4[/tex] is 6 : 4 or 3 : 2
As per the Equation
6 moles of water produces 4 moles of Phosphoric acid
So let us convert
12.81 moles [tex]H_2 O[/tex] to moles [tex]H_3 PO_4[/tex] by using mole ratio and then to mass [tex]H_3 PO_4[/tex] by multiplying with molar mass [tex]H_3 PO_4[/tex]
[tex]12.82 mol H_2 O \times \frac {(4mol H_3 PO_4)}{(6molH_2 O)} \times \frac {(98gH_3 PO_4)}{(1mol H_3 PO_4 )}[/tex]
[tex]=837g H_3 PO_4[/tex] is produced
Please note:
Molar mass is the mass of 1 mole of the substance and its unit is g/mol
We find the molar mass by adding the atomic mass of the atoms present in it.
For example [tex]H_3 PO_4[/tex] contains 3 atoms of H, 1 atom of P and 4 atoms of O
So the molar mass
[tex]=(3\times1.008)+(31\times1)+(4\times16)=98 g/mol[/tex]
Match each vocabulary word with its definition. match the items in the left column to the items in the right column. 1 . periodicity unlike charges 2 . shell like charges 3 . nonmetals charge on a neutron 4 . noble gases location of an electron 5 . neutral high ionization energies 6 . repel repeating nature of atomic structure 7 . attract least reactive family
Answer:
Explanation:
1 . periodicity
repeating nature of atomic structure
Periodicity is the repeating nature or trend of an atom on the periodic table. Periodicity can be inform of ionization energy, atomic radius, nuclear charge etc.
2 . shell
location of an electron
Electrons are located in shells. It is where electrons are found in an atom and the probability of finding electrons there are high
3 . non-metals
high ionization energies
Non-metals have high ionization energies which is the energy required to remove a loosely bonded electron in an atom. Metals have low ionization energies.
4 . noble gases
least reactive family
Noble gases are called inert gases. They have complete electronic configuration and this makes them unreactive.
5 . neutral
charge on a neutron
Neutrons have no charge on them. They are subatomic particles found in the nucleus alongside the protons
6 . repel
like charges
Like charges repel one another
7 . attract
unlike charges
Unlike charges attracts. Positve attracts negative charges.
Answer:
sorry in need question for points
Explanation:
1 . periodicity
repeating nature of atomic structure
Periodicity is the repeating nature or trend of an atom on the periodic table. Periodicity can be inform of ionization energy, atomic radius, nuclear charge etc.
2 . shell
location of an electron
Electrons are located in shells. It is where electrons are found in an atom and the probability of finding electrons there are high
3 . non-metals
high ionization energies
Non-metals have high ionization energies which is the energy required to remove a loosely bonded electron in an atom. Metals have low ionization energies.
4 . noble gases
least reactive family
Noble gases are called inert gases. They have complete electronic configuration and this makes them unreactive.
5 . neutral
charge on a neutron
Neutrons have no charge on them. They are subatomic particles found in the nucleus alongside the protons
6 . repel
like charges
Like charges repel one another
7 . attract
unlike charges
Unlike charges attracts. Positve attracts negative charges.
How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density of water is 1.0 g/mL.) Express your answer using two significant figures.
Approximately 88.6 grams of ice would need to melt to lower the temperature of the water from 26 ∘C to 6 ∘C.
Explanation:To determine the amount of ice that would need to melt to lower the temperature of the water, we can use the equation Q = mLf, where Q is the amount of heat transferred, m is the mass, and Lf is the latent heat of fusion. In this case, the water needs to be cooled from 26 ∘C to 6 ∘C, which is a decrease of 20 degrees. The specific heat capacity of water is 4.184 J/g °C, and the density of water is 1.0 g/mL. So, for the water to cool down, it would lose 4.184 J/g °C * 353 mL * 20 °C = 2.96 x 104 J of heat. To convert this to grams of ice, we can use the equation m = Q / Lf. The latent heat of fusion for water is 334 J/g. Substituting the values, we get m = 2.96 x 104 J / 334 J/g = 88.6 g. Therefore, approximately 88.6 grams of ice would need to melt to lower the temperature of the water from 26 ∘C to 6 ∘C.
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Approximately 3.44 grams of ice would need to melt to lower the temperature of 353 mL of water from 26 °C to 6 °C.
Explanation:To calculate the amount of ice that would need to melt to lower the temperature of water, we need to use the formula Q = mL. The heat transferred to melt the ice is equal to the mass of ice multiplied by the latent heat of fusion. In this case, the initial temperature of the water is 26 °C and the final temperature is 6 °C. We need to find the mass of ice in grams, so we can use the density of water, which is 1.0 g/mL. The volume of water is given as 353 mL, so the mass of water is 353 g. Since the density of water is the same as the density of ice, we can assume that the volume of ice is also 353 mL. Therefore, the mass of ice is also 353 g. We can now calculate the amount of ice that would need to melt.
Q = mL,
mL = Q / Lf,
m = (Q / Lf) / 1000
where Q is the amount of energy absorbed or released during a phase change, Lf is the latent heat of fusion (334 J/g), and m is the mass of ice.
Substituting the values into the equation, we get:
m = (Q / Lf) / 1000 = (1.15 × 10^6 J) / (334 J/g) / 1000 = 3.44 g.
Therefore, approximately 3.44 grams of ice would need to melt to lower the temperature of 353 mL of water from 26°C to 6 °C.
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Horizontal rows of the periodic table are called
Horizontal rows of the periodic table are called periods.
Periods are the horizontal rows of the periodic table that correspond to successive layers of electron shells surrounding the atomic nucleus. The atomic number rises as one moves over a period from left to right, adding more protons and electrons in the process.
The elements' characteristics gradually change as a result of this process, changing in reactivity and with an increase in atomic radius. While all elements in the same period have the same number of electron shells, the amounts of valence electrons in each element cause the elements' chemical characteristics to vary.
A mineral is observed to become soft and crumbly and develop a reddish color as it undergoes weathering, what process has most likely taken place?
Answer:
Corrosion
Explanation:
The corrosion is a natural process that occurs because of the influence of the water and wind. It is part of the processes of erosion. It manages to soften the minerals, make them less stable and more crumbly. It can easily be noticed as it gives the mineral reddish, or rather dark orange color. This process occurs where the humidity is high, as the water plays a crucial role for this process to occur. Apart from occurring in nature, this process is also a big problem with the refined metals, as it gradually destroys them, and taking in consideration that the metals are included in all sorts of objects, it can make a lot of damage.
Please answer these questions fast! Im on a time crunch right now
Answer:
1st image, 17.79 g Mg(NO3)2
2nd image 1.47 mol NH3
3rd image Molar Mass Al(OH)3 = 78 g/mol
Explanation:
1st image
Molar Mass Mg(NO3)2 = 148. 31 g/mol
0,12 mol Mg(NO3)2 x (148, 31 g Mg(NO3)2 / 1 mol Mg(NO3)2) = 17.79 gMg(NO3)2
2nd image
Molar Mass NH3 = 17g/mol
25 g NH3 x ( 1 mol NH3/ 17 g NH3) = 1.47 mol NH3
3rd image
Molar Mass Al(OH)3 = 1 Al *(27) + 3 O * (16) + 3 H * (1) = 78 g/mol
The amount of gas that occupies 60.82 l at 31.0 °c and 367 mm hg is ________ mol.
Answer:
1.17mole
Explanation:
Given parameters:
Volume of gas = 60.82L or 60.82dm³
Temperature of the gas = 31.0°C or 304K
Pressure on the gas = 367mmHg
To make calculation easier we convert the pressure from mmHg to atm.
1 atm = 760mmHg
367mmHg gives [tex]\frac{367}{760}[/tex]atm = 0.48atm
Solution
Assuming ideality, we can find the number of moles of the gas used. The ideal gas law combines Boyle's law, Charles's Law and Avogadro's law. It is expressed below:
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant given as 0.082atmdm³mol⁻¹K⁻¹
T is the temperature in kelvin
The unknown is n, which is the number of moles.
Making n the subject of the formula, we have ;
n = [tex]\frac{PV}{RT}[/tex]
n = [tex]\frac{0.48 x 60.82}{0.082x304}[/tex]= 1.17mole
The question is asking to identify the number of moles in a gas given certain conditions using the Ideal Gas Law (PV=nRT). We do this by first converting all units to the necessary format and then substituting them into the rearranged Ideal Gas Law equation to solve for n (moles).
Explanation:The student's question is basically asking about the amount of moles in a gas under specified conditions. We can decipher this answer through the Ideal Gas Law equation which states that PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. To apply the equation, we first need to convert the given conditions to the correct units compatible with R (Ideal Gas Constant) which is typically 0.0821 L·atm/K·mol.
First, volume (V) is already given in liters (60.82 L)Second, we convert temperature (T) from Celsius to Kelvin by adding 273.15 to the stated degree Celsius, so 31.0 °c becomes 304.15 K Third, we need to convert pressure (P) from mmHg to atm. The conversion factor is approximately 1 atm = 760 mmHg, so pressure becomes approximately 0.483 atmOnce we have these conditions correctly converted, we substitute them into the Ideal Gas Law. Rearranging the equation to solve for n (moles), the equation becomes n = PV/RT. Thus, substituting our numbers in gives n = (0.483 atm*60.82 L)/ (0.0821 L·atm/K·mol * 304.15 K).
Simplifying this will give the answer for the amount of moles.
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Write the equilibrium expression for this reaction: (C19H8Br4O5S)H– ↔ (C19H8Br4O5S)2– + H+
Answer:
Keq = [ (C₁₉ H₈ Br₄ O₅ S)²⁻ ] [ H⁺ ] / [ (C₁₉ H₈ Br₄ O₅ S) H⁻ ]Explanation:
The equilibrium expression for the general equilibrium represented by the equation:
AB ⇄ A ⁺ + B⁻,is:
Keq = [A⁺] [B⁻] / [AB]Then, following that sketch, you obtain:
Reaction:(C₁₉ H₈ Br₄ O₅ S) H⁻ ⇄ (C₁₉ H₈ Br₄ O₅ S)²⁻ + H⁺
Equilibrium expression:Keq = [ (C₁₉ H₈ Br₄ O₅ S)²⁻ ] [ H⁺ ] / [ (C₁₉ H₈ Br₄ O₅ S) H⁻ ]
The square brackets are used to mean molar concentration (molariy).
The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of stock solution. Instead, it is done by a series of dilutions. A sample of 0.6597 g of KMnO4 was dissolved in water and made up to the volume in a 500.0−mL volumetric flask. A 2.000−mL sample of this solution was transferred to a 1000−mL volumetric flask and diluted to the mark with water. Next, 10.00 mL of the diluted solution was transferred to a 250.0−mL flask and diluted to the mark with water. (a) Calculate the concentration (in molarity) of the final solution. Enter your answer in scientific notation.
Answer: [tex]6.4\times 10^{-7}M[/tex].
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.6597g}{158g/mol}=0.004mole[/tex]
[tex]V_s[/tex] = volume of solution in ml = 500 ml
[tex]Molarity=\frac{0.004\times 1000}{500}=0.008M[/tex]
According to the dilution law:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = 0.008 M
[tex]V_1[/tex] = volume of stock solution = 2 ml
[tex]M_2[/tex] = molarity of resulting solution = ?
[tex]V_2[/tex] = volume of resulting solution = 1000 ml
[tex]0.008\times 2 ml=M_2\times 1000ml[/tex]
[tex]M_2=0.000016M[/tex]
According to the dilution law:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = 0.000016 M
[tex]V_1[/tex] = volume of stock solution = 10 ml
[tex]M_2[/tex] = molarity of resulting solution = ?
[tex]V_2[/tex] = volume of resulting solution = 250 ml
[tex]0.000016\times 10 ml=M_2\times 250ml[/tex]
[tex]M_2=0.00000064M=6.4\times 10^{-7}M[/tex]
Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
Therefore, the concentration of of the final solution is [tex]6.4\times 10^{-7}M[/tex].
A solution is prepared by dissolving 0.6597 g of KMnO₄ in a 500.0 mL volumetric flask and then is subjected to 2 successive dilutions. The final molarity of the solution is 6.680 × 10⁻⁷ M.
A solution is prepared by dissolving 0.6597 g of KMnO₄ in a 500.0 mL volumetric flask and then is subjected to 2 successive dilutions.
The initial molarity of the solution is:
[tex]M = \frac{mass\ solute }{molar\ mass\ solute \times liters\ solution} = \frac{0.6597g}{158.03g/mol \times 0.5000L } = 8.349 \times 10^{-3} M[/tex]
First dilutionA 2.000−mL (V₁) sample of this solution was transferred to a 1000−mL (V₂) volumetric flask and diluted to the mark with water. We can calculate the final concentration using the dilution rule.
[tex]C_2 = \frac{C_1 \times V_1 }{V_2} = \frac{(8.349 \times 10^{-3}M )\times 2.000mL }{1000mL} = 1.670 \times 10^{-5}M[/tex]
Second dilution10.00 mL (V₁) of the diluted solution was transferred to a 250.0−mL (V₂) flask and diluted to the mark with water. We can calculate the final concentration using the dilution rule.
[tex]C_2 = \frac{C_1 \times V_1 }{V_2} = \frac{(1.670 \times 10^{-5}M )\times 10.00mL }{250.0mL} = 6.680 \times 10^{-7}M[/tex]
A solution is prepared by dissolving 0.6597 g of KMnO₄ in a 500.0 mL volumetric flask and then is subjected to 2 successive dilutions. The final molarity of the solution is 6.680 × 10⁻⁷ M.
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Hydrogen gas can be prepared by reacting metallic sodium with water: 2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq). If the price of sodium were 165/kg, what would be the cost of producing 1.00 mol of hydrogen. Ignore the cost of the water.
Answer: cost of Na = 7.59
Explanation:
From the given balanced equation , 2 mol Na/2x23 g produced 1 mol of H2 gas
And cost of Na is 165 for per Kg or 1000 g Na
Then cost of Na for 2x23=46 g Na
Cost = 46x165/1000= 7.59
Answer:
$7.59
Explanation:
From the balanced reaction equation, 2moles of sodium is required to produce 1.0 mole of hydrogen.
Now, the mass of 2 moles of sodium is 23.0×2 = 46 of sodium.
The mass of sodium required in kilograms is 46/1000= 0.046Kg of sodium
If 1 Kg of sodium cost $165
0.046Kg of sodium will cost 0.046×165 = $7.59
Therefore it will cost $7.59 to produce 1.0 moles of hydrogen gas according to the equation.
The standard enthalpy change ΔH o rxn for the thermal decomposition of silver nitrate according to the following equation is 78.67 kJ: AgNO3(s) → AgNO2(s) + 1 2 O2(g) The standard enthalpy of formation of AgNO3(s) is −123.02 kJ/mol. Calculate the standard enthalpy of formation of AgNO2(s).
Answer: The enthalpy of the formation of [tex]AgNO_2(s)[/tex] is coming out to be -44.35 kJ/mol.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
For the given chemical reaction:
[tex]AgNO_3(s)\rightarrow AgNO_2(s)+\frac{1}{2}O_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(AgNO_2)})+(\frac{1}{2}\times \Delta H^o_f_{(O_2)})]-[(1\times \Delta H^o_f_{(AgNO_3)})][/tex]
We are given:
[tex]\Delta H^o_f_{(AgNO_3)}=-123.02kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_{rxn}=78.67kJ[/tex]
Putting values in above equation, we get:
[tex]78.67=[(1\times \Delta H^o_f_{(AgNO_2)})+(\frac{1}{2}\times 0)]-[1\times (-123.0))]\\\\\Delta H^o_f_{(AgNO_2)}=-44.35kJ/mol[/tex]
Hence, the enthalpy of the formation of [tex]AgNO_2(s)[/tex] is coming out to be -44.35 kJ/mol.
The standard enthalpy of formation AgNO2(s) is calculated from the standard enthalpy change for the thermal decomposition of silver nitrate and the standard enthalpy of formation of AgNO3(s). It is found to be -44.35 kJ/mol.
Explanation:This question is based on the calculation of the standard enthalpy of formation for AgNO2(s). The standard enthalpy change ΔH for the thermal decomposition of silver nitrate is given as 78.67 kJ according to the following equation: AgNO3(s) → AgNO2(s) + 1 2 O2(g). Additionally, we know that the standard enthalpy of formation of AgNO3(s) is −123.02 kJ/mol.
The standard enthalpy of formation for AgNO2 can be calculated using the formula ΔHf [AgNO2] = ΔH [reaction] + ΔHf [AgNO3]. Substituting the given values, we get ΔHf [AgNO2] = 78.67 kJ/mol + (-123.02 kJ/mol) = -44.35 kJ/mol. Therefore, the standard enthalpy of formation of AgNO2(s) is -44.35 kJ/mol.
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The boiling point of chlorine is −34 °C. Which of the following best predicts the boiling point of iodine?
a) Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.
b) Lower than −34 °C because chlorine is more polar than iodine on account of its higher electronegativity.
c) Higher than −34 °C because dipole-dipole interactions in iodine are stronger than dispersion forces in chlorine.
d) Lower than −34 °C because permanent dipoles created in chlorine are stronger than temporary dipoles in iodine.
Answer:
a) Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.
Explanation:
Both chlorine and Iodine are halogens. They belong to the 7th group on the periodic table.
To examine the trend of boiling point in this group we must consider the nature of the intermolecular bonds that would be formed by the molecules of these elements. We know that London dispersion forces, a type of van der Waals attraction would be more prevalent. This is so because the molecules here would be non-polar and the uneven distribution of the constantly moving electrons would initiate the intermolecular bonding.
The uneven charge distribution leads to the formation of a temporary dipole.
This makes boiling point increase down the group because more electrons becomes involved and the bond becomes stronger.
Answer:
Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.
Explanation:
A solution has a pH of 4.20. Using the relationship between pH and pOH, what is the concentration of OH?? A. 9.9 × 10-1 M B. 6.2 × 10-1 M C. 6.3 × 10-5 M D. 6.7 × 10-6 M E. 1.6 × 10-10M
The correct answer is option E.
Explanation:
The pH of the solution is defined as negative logarithm of hydrogen ions concentration in a solution. Mathematically written as:
[tex]pH=-\log[H^+][/tex]
The pOH of the solution is defined as negative logarithm of hydroxide ions concentration in a solution. Mathematically written as:
[tex]pOH=-\log[OH^-][/tex]
The sum of ph and pOH is equal to 14.
pH + pOH = 14
A solution has a pH of 4.20.
[tex]pOH=14-pH=14-4.20=9.8[/tex]
[tex]pOH=9.8=-\log[OH^-][/tex]
[tex][OH^-]=1.58\times 10^{-10} M\approx 6.0\times 10^{-10} M[/tex]
Hence, the correct answer is option E.
Answer:
E. 1.6 x 10 -10 M is the correct answer
Determine whether each statement describes the solid state, liquid state, or gaseous state.
1. This state is characterized by the lowest density of the three states.
2. This state is characterized by an indefinite shape and high density.
3. Temperature changes significantly influence the volume of this state.
4. Pressure changes influence the volume of this state more than the other two states.
5. In this state, constituent particles are less free to move around than in other states.
Answer:
1. Gas
2. Liquid
3. Solid
4. Gas
5. Solid
Hope this helps!
The gaseous state is characterized by the lowest density and is influenced by temperature and pressure changes. The liquid state has an indefinite shape and a higher density. The solid state has particles that are less free to move around than in the other states.
Explanation:1. This state is characterized by the lowest density of the three states. This describes the gaseous state, as gas particles are spread out and not closely packed together.
2. This state is characterized by an indefinite shape and high density. This refers to the liquid state, which adapts its shape to fit its container but has particles that are still relatively close together, contributing to a higher density than gases.
3. Temperature changes significantly influence the volume of this state. This is typical of the gaseous state, as gases expand or contract significantly with temperature changes according to the Ideal Gas Law.
4. Pressure changes influence the volume of this state more than the other two states. This also pertains to the gaseous state, where pressure changes can lead to notable volume changes, as predicted by the Ideal Gas Law.
5. In this state, constituent particles are less free to move around than in other states. This describes the solid state, where particles are in fixed positions and can only vibrate in place.
Two different compounds are formed by the elements nitrogen and oxygen. The first compound, compound P1, contains 64.17% by mass nitrogen and 35.73% by mass oxygen. The second compound, compound P2, contains 47.23% by mass nitrogen and 52.79% by mass oxygen. What is the ratio of nitrogen to a fixed mass of oxygen for the two compounds?
Answer:
Explanation:
To solve the problem, we must know the kind of compounds we are dealing with.
For the first compound, P1 and second compound P2:
N O N O
Mass percent 64.17 35.73 47.23 52.79
Atomic mass 14 16 14 16
Number of
moles 64.17/14 35.73/16 47.23/14 52.79/16
4.58 2.23 3.37 3.30
Simplest
ratio 4.58/2.23 2.23/2.23 3.37/3.30 3.3/3.3
2 1 1 1
P1 compound is N₂O
P2 compound is NO
These are the compounds,
In N₂O = 28:16
NO = 14:16
This is the ratio of nitrogen to a fixed mass of oxygen for the two compounds.
Which one of the following substances has three unpaired d electrons? Which one of the following substances has three unpaired d electrons? [Cu(NH3)4]2+ [V(H2O)6]4+ [Ag(NH3)2]+ [Zn(NH3)4]2+ [Cr(CN)6]3-
Answer:
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Explanation:
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In which of the following reactions will Kc = Kp? In which of the following reactions will Kc = Kp?
(A) SO3(g) + NO(g) ↔ SO2(g) + NO2(g)
(B) 2 N2(g) + O2(g) ↔ 2 N2O(g)
(C) 4 NH3(g) + 3 O2(g) ↔ 2 N2(g) + 6 H2O(g)
(D) 2 SO2(g) + O2(g) ↔ 2 SO3(g)
(E) None of the above reactions have Kc = Kp.
Answer: Option A
Explanation:
Since we have a relationship between KC and KP as given below
KP=KC(RT)^dn
dn= difference in the no of gaseous products and Reactants
For A,dn= 1+1-(1+1)= 2-2=0
KP=KC(RT)^0
KP=kc(1) since a^0=1
Hence KP=KC
For B,dn=2-(2+1)=2-3=-1, hence KP not equal to KC
For C, dn=2+6-(4+3)=8-7=1
For D,dn=2-3=-1, KP is not equal to KC
So option A is correct answer
The equilibrium gas constants of the gaseous mixture are Kc and Kp. Equilibrium constants are dependent on moles and partial pressure. Kc = Kp in [tex]\rm SO_{3}(g) + NO(g) \leftrightarrow SO_{2}(g) + NO_{2}(g).[/tex]
What is the relationship between Kc and Kp?Kc is the equilibrium constant of the moles while Kp is of the partial pressure, the relation between these two are:
[tex]\rm Kp=Kc(RT)^{dn}[/tex]
Here dn is the difference of the amount or the number of the products and the reactants of the reaction.
The value of dn for reaction A is 1+1-(1+1)=0, since the value of the power is zero Kp will be equivalent to Kc. For B, C and D reactions the value of dn are -1, 1 and -1 respectively hence Kp is not equal to Kc.
Therefore, the reaction in option A. [tex]\rm SO_{3}(g) + NO(g) \leftrightarrow SO_{2}(g) + NO_{2}(g)[/tex] will have an equal value of Kp and Kc.
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The probability that a continuous random variable takes any specific value
a. is equal to zero
b. is at least 0.5
c. depends on the probability density function
d. is very close to 1.0
Answer:
a. It is equal to zero
Explanation:
As the possible results of a continuous random variable are infinite, it is most common to talk about probabilities on intervals; A singular and particular value is so difficult to obtain in an experiment because there are infinite possibilities, and a specific value is just a very little part of the possibilities.
That is the reason why, as a definition, that probability is always zero when you have a continuous random variable.
The probability that a continuous random variable takes a specific value is zero; instead, probabilities are defined over intervals as found using the probability density function.
Explanation:The probability that a continuous random variable takes any specific value is equal to zero. This is because every point in a continuous space has an infinitely small probability, and hence, when we speak about probabilities for continuous random variables, we talk in terms of intervals rather than specific values. This is reflected in the probability density function (pdf), which provides the likelihood over an interval. Calculating the probability for an event involving a continuous random variable involves finding the area under the pdf curve between two points that define the range of interest.
As an example, if we have P(x < 5) = 0.35, it follows that P(x > 5) = 1 - P(x < 5) = 0.65, given that the total area under the pdf curve sums up to 1. This principle illustrates how continuous probability distributions work and why probabilities for specific values are not feasible.
Calculate the total mass of the protons and electrons in 19 9F. Use 1.007825 amu as the mass of 11H (mass of a proton and an electron). Express your answer in atomic mass units to 6 significant figures.
Explanation:
Atomic mass means the sum of total number of protons and neutrons present in an atom.
When an atom is neutral then number of protons equal the number of electrons.
Mass of a proton equals to 1.007276 u, mass of neutron equals to 1.008664 u.
For example, in a [tex]^{19}_{9}F[/tex] atom there are total 9 protons and number of neutrons is 19 - 9 = 10.
Since, it is a neutral atom so number of electrons will also be 9.
So, total mass will be calculated as follows.
Total mass = 9 (mass of electron + mass of proton)
= 9 (1.007825 u)
= 9.07043 u
Therefore, we can conclude that total mass of protons and electrons in [tex]^{19}_{9}F[/tex] is 9.07043 u.
The total mass of the protons and electrons in 19 9F is approximately 9.07065 amu. This calculation is made by multiplying the number of each particle by their respective masses and summing the results.
Explanation:In order to calculate the total mass of the protons and electrons in 19 9F, you first need to understand that the number '9' represents the atomic number, which is the number of protons in the atom, and hence, in a neutral atom, the number of electrons as well. The mass of a proton is roughly 1.0073 amu and the mass of an electron is approximately 0.00055 amu. With 9 protons and 9 electrons, the total mass would be (9 protons * 1.0073 amu/proton)+(9 electrons * 0.00055 amu/electron).
So, the mass of the protons and electrons is equal to 9.0657 amu for the protons and 0.00495 amu for the electrons. Adding these together gives a total mass of approx 9.07065 amu.
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The particles of a substance are arranged in a definite geometric pattern and are constantly vibrating. This substance can be in
1, the souls phase
2, the liquid phase,only
3, either the solid or the liquid phase
4, neither the liquid nor the solid phase
Answer:
1. Solid phase.
Explanation:
The molecules in a solid are held in a definite pattern which is not true of a liquid or gas.
Answer:
Solid State
Explanation:
in the solid state of matter substances have the particles arranged in a definite geometric pattern and are constantly vibrating, the main difference is the arrangement of the particles in a definite geometric pattern, compared to the liquid state in which particles take the shape of the conteiner where they are in.
What processes can change sedimentary rock into igneous rock?
A. High pressure and deposition
B. Lithification and cooling
C. Uplift and erosion
D. Heating and cooling
Heating and cooling are the processes can change sedimentary rock into igneous rock.
What processes can change sedimentary rock into igneous rock?
When Sedimentary rocks are heated with tremendous heat and pressure, it will melt and be back again to magma. After some time it will cool and harden and will become Igneous rocks.
So we can conclude that Heating and cooling are the processes can change sedimentary rock into igneous rock.
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Answer:
Erosion and lithification
Explanation:
One disadvantage of vacuum distillation is it is inherently less efficient than fractional distillation at atmospheric pressure.
a. True
b. False
Answer:
One disadvantage of vacuum distillation is it is inherently less efficient than fractional distillation at atmospheric pressure. :True -a.
You are performing an experiment in a lab to attempt a new method of producing pure elements from compounds. The only problem is that you do not know what element will form. By your previous calculations you know that you will have 3 moles of product. When it is complete, you weigh it and determine you have 209.1 grams. What element have you produced?
You need to calculate the atomic weight of that element.
number of moles = mass / atomic weight
atomic weight = mass / number of moles
atomic weight = 209.1 / 3 = 69.7 g/mol
The element produced will be gallium, Ga.
Answer:
The element produced is gallium, Ga
Explanation:
Given:
Moles of the product = 3 moles
Weight of the product = 209.1 g
To determine:
The identity of the product formed i.e. the unknown element
Calculation:
The identity of element can be deduce from its atomic weight and comparing the calculated weight to that of the elements in the periodic table.
[tex]Number\ of\ moles = \frac{mass\ of\ the\ element}{atomic\ weight}[/tex]
[tex]Atomic\ weight = \frac{mass\ of\ the\ element}{number\ of\ moles}[/tex]
In this case:
[tex]Atomic\ weight = \frac{209.1\ g}{3\ moles} = 69.7\ g/mol[/tex]
From the periodic table, the element with an atomic mass = 69.7 g/mol is gallium, Ga
Identify the balanced chemical equation that represents a single displacement reaction. CF4 2Br2 ⟶ CBr4 2F2 3H2SO4 2Al ⟶ Al2(SO4)3 3H2 3H2SO4 2Al ⟶ Al2(SO4)3 H2 CF4 Br2 ⟶ ⟶ CBr4 F2
Answer:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
Explanation:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
In this type of reaction, one substance is replacing another:
A + BC → AC + B
In a single displacement reaction, atoms replace one another based on the activity series. Elements that are higher in the activity series. Also, if the element that is to replace the other in a compound is more reactive the reaction will occur. If it is less reactive, there will be no reation.
In the first equation, fluorine is more reactive than bromine. Therefore, bromine cannot replace bromine.
In the second equation, the displacement is between hydrogen and aluminium. Hydrogen is lower in the activity series, this implies that aluminum will replace it.
The balanced equation for a single displacement reaction is 3H2SO4 + 2Al
ightarrow Al2(SO4)3 + 3H2, where aluminum displaces hydrogen in sulfuric acid to form aluminum sulfate and hydrogen gas.
To identify the balanced chemical equation that represents a single displacement reaction, we can look at the provided options. A single displacement reaction is a type of chemical reaction where one element is replaced by another in a compound. We can consider the following valid general form for a single displacement reaction: A + BC
ightarrow B + AC, where A replaces B in the compound BC.
Now, the correct balanced equation among the provided examples that represents a single displacement reaction is:
3H2SO4 + 2AlThis equation follows the pattern of a single displacement reaction, where aluminum (Al) displaces the hydrogen (H) in sulfuric acid (H2SO4) to form aluminum sulfate (Al2(SO4)3) and hydrogen gas (H2). The equation is also balanced, with equal numbers of each type of atom on both sides of the equation.
Sodium bicarbonate is sometimes used to neutralize acid spills. If 2.0 L of 1.0 M H2SO4 is spilled, what mass of sodium bicarbonate (NaHCO3) is required to neutralize all the H2SO4?
Answer:
168.0 g
Explanation:
First thing, write a balanced chemical equation:
[tex]H_{2}SO_{4}+ 2NaHCO_{3} ----> Na_{2}SO_{4} + 2H_{2}O+2CO_{2}[/tex]
n(H2SO4) = concentration * volume
= 1.0 M * 2.0 L
= 2.0 mol
According to the balanced equation, 1 mol of acid requires 2 mol of sodium bicarbonate. This means that 2 mol of acid requires 2 mol of sodium bicarbonate. What mass of sodium bicarbonate is this?
mass (NaHCO3) = number of moles * molar mass
= 2.0 mol * 84.0065 g/mol
= 168.0 g
To neutralize the 1.0 M H2SO4, you will need 336.04 g of sodium bicarbonate (NaHCO3).
Explanation:To neutralize the 1.0 M H2SO4, we can use the balanced chemical equation for the reaction between sodium bicarbonate (NaHCO3) and sulfuric acid (H2SO4):
2 NaHCO3 + H2SO4 → Na2SO4 + 2 H2O + 2 CO2
From the equation, we can see that 2 moles of NaHCO3 are required to neutralize 1 mole of H2SO4. Therefore, we can determine the mass of NaHCO3 required using the molar mass and stoichiometry:
Molar mass of NaHCO3 = 22.99 g/mol + 1.01 g/mol + 3(16.00 g/mol) = 84.01 g/mol1.0 M H2SO4 means it contains 1 mole of H2SO4 per liter of solution2.0 L of 1.0 M H2SO4 contains 2 moles of H2SO42 moles of H2SO4 requires 4 moles of NaHCO3 (according to the balanced equation)The mass of NaHCO3 required = 4 moles × 84.01 g/mol = 336.04 gTherefore, the mass of sodium bicarbonate required to neutralize the H2SO4 is 336.04 g.
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