Answer:
less becuase that moisture adds weight
Explanation:
What volume of 6.00 M hydrochloric acid is needed to prepare 500 mL of 0.100 M solution?
Answer:
8.33mL or .0083L
Explanation:
Use m1 * V1 = m2 * V2
6.00M(x) = 0.100M(500mL)
solve for x
x= (.1 * 500) / 6
x=8.333 mL
For the following reaction, 142 grams of silver nitrate are allowed to react with 22.3 grams of copper . silver nitrate(aq) copper(s) copper(II) nitrate(aq) silver(s) What is the maximum amount of copper(II) nitrate that can be formed
Answer:
even I have the same dought
Calculate the change in entropy when 1.00 kgkg of water at 100∘C∘C is vaporized and converted to steam at 100∘C∘C. Assume that the heat of vaporization of water is 2256×103J/kg2256×103J/kg. Express your answer in joules per kelvin.
Answer:
[tex]\Delta S=6045.8\frac{J}{K}[/tex]
Explanation:
Hello,
In this case, we can compute the change in the entropy for vaporization processes in term of the enthalpy of vaporization as shown below:
[tex]\Delta S=\frac{m*\Delta H}{T}[/tex]
Whereas the temperature is in Kelvins. In such a way, the entropy results:
[tex]\Delta S=\frac{1.00kg*2256x10^3\frac{J}{kg} }{(100+273.15)K}\\\\\Delta S=6045.8\frac{J}{K}[/tex]
Best regards.
In the following chemical reaction, what are the products? 6H₂O+6CO₂→C₆H₁₂O₂+6O₂ Options: A) C₆H₁₂O₂+6O₂ B) 6H₂O+6CO₂ C) 6CO₂+6O₂ D) 6H₂O+C₆H₁₂O₆
Answer:
A
Explanation:
The products are the stuff on the right side of the arrow (yield sign). In this case that would be C₆H₁₂O₂ + 6O₂.
Each unknown mixture contains 5 metal constituents. Select the 5 metal ions that you have identified as being present in your mixture. Please double check your selections before you hit the submit button. a. Ca b. Co c. Cr d. Fe e. K f. Mn g. Zn
Explanation:
A metal ion is a type of atom compound that has an electric charge.
Such atoms willingly lose electrons in order to build positive ions called cations. The selected Ions are :
[tex]1. Mn^2^+\\\ 2. Ca^2^+\\\ 3. Co^2^+\\\ 4. Fe^2^-\\\ 5. K^+[/tex]
Choose the situation below that would result in an endothermic ΔHsolution.
a) When |ΔHsolute| > |ΔHhydration|
b) When |ΔHsolute| is close to |ΔHhydration|
c) When |ΔHsolute| < |ΔHhydration|
d) When |ΔHsolvent| >> |ΔHsolute|
e) There isn't enough information to determine.
Answer:
Option A - When |ΔHsolute| > |ΔHhydration|
Explanation:
A solution is defined as a homogeneous mixture of 2 or more substances that can either be in the gas phase, liquid phase, solid phase.
The enthalpy of solution can either be positive (endothermic) or negative (exothermic).
Now, we know that enthalpy is amount of heat released or absorbed during the dissolving process at constant pressure.
Now, the first step in thus process involves breaking up of the solute. This involves breaking up all the intermolecular forces holding the solute together. This means that the solute molecules are separate from each other and the process is always endothermic because it requires energy to break interaction. Thus;
The enthalpy ΔH1 > 0.
Thus, the enthalpy of the solute has to be greater than the enthalpy of hydration.
An endothermic ΔHsolution occurs when |ΔH solute| < |ΔH hydration|.
A substance dissolves in water when the solute - solvent interaction exceeds the solute - solute solute interaction. The energy required to break the bonds between solutes is the ΔHsolute and the energy released when solute - solvent interaction take place is called the ΔHhydration.
We know that when |ΔH solute| < |ΔH hydration|, energy is required to break up the solute - solute interaction and ΔHsolution is endothermic.
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2) Which of the following elements is not part of the atom?
O Nucleolus
O Core
OElectron
O Proton
Answer:
the nucelolo is not part of the atom
Explanation:
The nucleolus is NOT a part of the atom. The nucleolus has its definition in biology as an element of the cell.
Let's remember that the atom is made up of three fundamental elements that are: electrons, protons and neutrons. Protons and neutrons are found in the nucleus while electrons are orbiting around the nucleus.
Answer:
core
Explanation:
there is no such thing as a core in an atom.
The middle is known as the nucleus
What is the change in energy, in kJ, when 45.3 grams of methanol, CH3OH, combusts? 2\text{CH}_3\text{OH}\left(l\right) + 3\text{O}_2\left(g\right)\rightarrow2\text{CO}_2\left(g\right)+4\text{H}_2\text{O}\left(g\right)\hskip .5in \Delta\text{H}=-726\text{ kJ}2 CH 3 OH ( l ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 4 H 2 O ( g ) Δ H = − 726 kJ Group of answer choices -513 kJ +2,050 kJ -1,030 kJ -2,050 kJ +513 kJ
Answer: -1,030 kJ
Explanation:
To calculate the number of moles we use the equation:
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar mass}}[/tex] .....(1)
Putting values in equation 1, we get:
[tex]\text{Moles of methanol}=\frac{45.3g}{32g/mol}=1.42mol[/tex]
The balanced chemical reaction is:
[tex]CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex] [tex]\Delta H=-726kJ[/tex]
Given :
Energy released when 1 mole of [tex]CH_3OH[/tex] combusts = 726k J
Thus Energy released when 1.42 moles of [tex]CH_3OH[/tex] combusts = [tex]\frac{726kJ}{1}\times 1.42=1030J[/tex]
Thus 1030 kJ of heat is released and as [tex]\Delta H[/tex] is negative for exothermic reaction, [tex]\Delta H=-1030kJ[/tex]
Turn on Write equation. What you see is an equation that shows the original uranium atom on the left. The boxes on the right represent the daughter product—the atom produced by radioactive decay—and the emitted alpha particle.
Answer:
Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.
Explanation:
Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:
[tex]_{92} ^{238} U_{}[/tex]→ [tex]_{90} ^{234} Th_{} + _{2} ^{4} He_{}[/tex]
I hope this explanation is clear and explanatory.
Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe: PbO(s) NaCl(aq) H2O(l) --> Pb(OH)Cl(s) NaOH(aq) How many grams of PbO and how many grams of NaCl would be required to produce 10.0 g of Pb(OH)Cl
Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
[tex]= \frac{Mass}{Molar\ mass}[/tex]
[tex]= \frac{10.0 g}{259.65g / mol}[/tex]
= 0.0385 mol
Mass of PbO needed is
[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.
If unknown to you, your pipet was incorrectly calibrated so that it transferred less than 10.00 mL of your solution, the density you calculated for the liquid would tend to be smaller or larger than the correct value. Explain.
Answer:
The density would be larger than the correct value.
Explanation:
First off, the realtionship between denisty and volume is given in the equation below;
Density = Mass / Volume
From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.
Assuming all thing's being normal;
Mass = 2g
Volume = 10ml
Density = 2 / 10 = 0.2 g/ml
Second case scenario;
'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"
Lets have a value of 8ml for our volume. Mass remains constant.
Density = 2 / 8 = 0.25 g/ml
The density would be larger than the correct value.
Answer: The density would be larger than the correct value.
First off, the relationship between density and volume is given by:
Density = Mass / Volume
From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.
Assuming all thing's being normal;
Mass = 2g
Volume = 10ml
Density = [tex]\frac{2}{10}=0.2[/tex] g/ml
Second case scenario;
'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"
Lets have a value of 8ml for our volume. Mass remains constant.
Density = [tex]\frac{2}{8}= 0.25[/tex] g/ml
The density would be larger than the correct value.
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What is the half-life for the first order decay of 14C according to the reaction, 146C — 147N +e- ?
The rate constant for the decay is 1.21 x10-4 year-1
Answer:
5727 years or 5730 (rounded to match 3 sig figs) whichever one your teacher prefers
Explanation:
First Order decay has a half life formula of Half Life = Ln (2) / k = 0.693/K
Half-life = 0.693/k = 0.693/1.21 x10-4 = 5727 years or 5730 (rounded to match 3 sig figs)
This should be correct because if you google the half-life of 14 C it is ~ 5700 years
A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is saying 91g I keep getting 175g. Can someone please explain?
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
What is the specific heat of a 85.01 g piece of an unknown metal that exhibits a 45.2°C temperature change upon absorbing 1870 J of heat?
Answer:
[tex]0.48~\frac{J}{g~^{\circ}C}[/tex]
Explanation:
In this question, we have to remember the relationship between Q (heat) and the specific heat (Cp) the change in temperature (ΔT), and the mass (m).
[tex]Q=m*Cp*ΔT[/tex]
The next step is to identify what values we have:
[tex]Q~=~1870~J[/tex]
[tex]m~=~85.01~g[/tex]
[tex]ΔT~=~45.2~^{\circ}C[/tex]
[tex]Cp~=~X[/tex]
Now, we can plug the values and solve for "Cp":
[tex]1870~J=~85.01~g~*Cp*45.2~^{\circ}C[/tex]
[tex]Cp=\frac{1870~J}{85.01~g~*45.2~^{\circ}C}[/tex]
[tex]Cp=0.48~\frac{J}{g~^{\circ}C}[/tex]
The unknow metal it has a specific value of [tex]0.48~\frac{J}{g~^{\circ}C}[/tex]
I hope it helps!
When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.430 g of a particular hydrocarbon was burned in air, 0.446 g of CO, 0.700 g of CO2, and 0.430 g of H2O were formed.
Required:
a. What is the empirical formula of the compound?
b. How many grams of O2 were used in the reaction?
c. How many grams would have been required for complete combustion?
Answer:
(a) The empirical formula of the compound is
m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).
(b) The grams of O2 that were used in the reaction is 1.146 g
(c) The amount of O2 that would have been required for complete combustion is 1.401 g.
Explanation:
a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)
(b) Using law of conservation of mass from above
m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)
m(O2) = 0.446 + 0.700 + 0.430 - 0.430
m(O2) = 1.146 g
The grams of O2 that were used in the reaction is 1.146 g
(c) for complete combustion, we need to oxidized CO to CO2
Then, 2CO +O2 = 2CO2
m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}
m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g
Note; Molar mass of O2 = 32, CO = 28
m(total)(O2) = m(O2) + m(add)(O2)
m(total)(O2) = 1.146 + 0.255 = 1.401 g
The amount of that grams would have been required for complete combustion is 1.401 g.
Note (add) and (total) were used subscript to "m"
A chemist measures the energy change Delta H during the following
2Fe2O3(s)->4FeO(s)+O2(g).
1) this reactions is: Endothermic or exothermic.
2) suppose 94.2g of Fe2O3 react. will any heat be relased or absorbed. yes absorbed. yes releases. no.
3) If you said heat will be released or absorbed in the second part of the question. calculate how much heat will be absored or released. be sure your answer has correct number of significant digits.
Answer: 1) Endothermic
2) Yes, absorbed.
3) 166.86 kJ will be absorbed.
Explanation:
1) To determine if a reaction is endothermic (heat is absorbed by the system) or exothermic (heat is released by the system), first calculate its change in Enthalpy, which is given by:
ΔH = [tex]H_{products} - H_{reagents}[/tex]
For the reaction 2Fe₂O₃(s) ⇒ 4FeO(s) + O₂(g):
Enthalpy of Reagent (Fe₂O₃(s))
Enthalpy of formation for Fe₂O₃(s) is - 822.2 kJ/mol
The reaction needs 2 mols of the molecule, so:
H = 2(-822.2)
H = - 1644.4
Enthalpy of Products (4FeO(s) + O₂(g))
Enthalpy of formation of O₂ is 0, because it is in its standard state.
Enthalpy of formation of FeO is - 272.04 kJ/mol
The reaction produces 4 mols of iron oxide, so:
H = 4(-272.04)
H = -1088.16
Change in Enthalpy:
ΔH = [tex]H_{products} - H_{reagents}[/tex]
ΔH = - 1088.16 - (-1644.4)
ΔH = + 556.2 kJ/mol
The change in enthalpy is positive, which means that the reaction is absorving heat. Then, the chemical reaction is Endothermic.
2) When Fe₂O₃(s) reacts, heat is absorbed because it is an endothermic reaction.
3) Calculate how many mols there is in 94.2 g of Fe₂O₃(s):
n = [tex]\frac{mass}{molar mass}[/tex]
n = [tex]\frac{94.2}{160}[/tex]
n = 0.6 mols
In the reaction, for 2 mols of Fe₂O₃(s), 556.2 kJ are absorbed. Then:
2 mols --------------- 556.2 kJ
0.6 mols ------------- x
x = [tex]\frac{0.6*556.2}{2}[/tex]
x = 167 kJ
It will be absorbed 167 kJ of energy, when 94.2 g of Fe₂O₃(s) reacts.
Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N 2O 4 and 45.0 g N 2H 4. Some possibly useful molar masses are as follows: N 2O 4 = 92.02 g/mol, N 2H 4 = 32.05 g/mol.
N 2O 4( l) + 2 N 2H 4( l) → 3 N 2( g) + 4 H 2O( g)
a) LR = N2O4, 45.7 g N2 formed
b) LR = N2O4, 105 g N2 formed
c) LR = N2H4, 13.3 g N2 formed
d) LR = N2H4, 59.0 g N2 formed
e) No LR, 45.0 g N2 formed
Answer:
Option A. LR = N2O4, 45.7g N2 formed
Explanation:
The balanced equation for the reaction is given below:
N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)
Next, we shall determine the masses of N2O4 and N2H4 that reacted and mass of N2 produced from the balanced equation. This is illustrated below:
Molar mass of N2O4 = 92.02 g/mol
Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02 g
Molar mass of N2H4 = 32.05 g/mol
Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g
Molar mass of N2 = 2x14.01 = 28.02g/mol
Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g
Summary:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4 to produce 84.06g of N2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4.
Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.
From the calculations made above, we can see that only 34.83g out 45g of N2H4 is required to react completely with 50g of N2O4.
Therefore, N2O4 is the limiting reactant and N2H4 is the excess reactant.
Finally, we shall determine the mass of N2 produced from the reaction.
In this case the limiting reactant will be used as it will produce the maximum yield of N2 since all of it is used up in the reaction.
The limiting reactant is N2O4 and the mass N2 produced can be obtained as illustrated below:
From the balanced equation above,
92.02g of N2O4 reacted to produce 84.06g of N2.
Therefore 50g of N2O4 will react to produce = (50 x 84.06)/92.02 = 45.7g of N2.
Therefore, 45.7g of N2 were produced from the reaction.
At the end of the day,
The limiting reactant is N2O4 and 45.7g of N2 were produced from the reaction.
Hypothesis 1: If you increase the
temperature of a reaction, then the reaction
rate will increase because particles experience
more collisions at higher temperatures.
To test the first hypothesis, you measured the
reaction rate for several different?
volumes.
temperatures.
densities.
particle sires.
Answer:is increase and more collisions :)
Explanation:
What is the pressure of 5.0 Mol nitrogen (N2) gas in a 2.0 L container at 268 K?
Answer:
pressure is = 54.9802atm
Explanation:
using ideal gas equation
PV=nRT
Which best describes the act of using senses or tools to gather information? creating a hypothesis making an observation summarizing the results recording the measurements
Answer:
B - Making an Observation
Explanation:
Making an observation best describes the act of using senses or tools to gather information. Therefore, option B is correct.
What are senses in the scientific method?The five senses—sight, taste, touch, hearing, and smell—gather data about our surroundings that the brain interprets. Based on prior experience (and subsequent learning), as well as by combining the data from each sensor, we make sense of this information.
Information gleaned from your five senses is referred to as an observation. These are smell, taste, touch, hearing, and sight. When you see a bird or hear it sing, you notice it.
The term observation, which is also used to sense five aspects of the world including vision, taste, touch, smell, and hearing, is used to describe utilizing the senses to examine the world, employing tools to take measurements, and looking at prior research findings.
Thus, option B is correct.
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Draw the structure of the organic product(s) of the Grignard reaction between dimethyl carbonate (CH3OCO2CH3) and excess phenylmagnesium bromide, followed by aqueous workup. You do not have to consider stereochemistry. If a compound is formed more than once, add another sketcher and draw it again. Alternatively, you may use the square brackets tool to add stoichiometries greater than one. Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner. Separate multiple products using the + sign from the dropdown menu.
Answer:
dimethoxy(phenyl)methanol
Explanation:
For this question, we have to remember the mechanism of the Grignard reaction. In this case, phenylmagnesium bromide is our nucleophile, a carbo-anion is produced (step 1). Then this carbo-anion can attack the carbonyl group in the dimethyl carbonate, the double bond is delocalized into the oxygen producing a negative charge (step 2). Finally, with the addition of the hydronium ion ([tex]H_3O^+[/tex]), the anion can be protonated to produce the alcohol (dimethoxy(phenyl)methanol) (step 3).
See figure 1
I hope it helps!
Description (with words) of water just above melting temperature. What intermolecular forces do you expect to find in water in liquid state
Answer:
intermolecular dipole-dipole hydrogen bonds
Explanation:
Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.
Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.
Write the empirical formula
Answer:
[tex]1) NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3) NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}[/tex]
Explanation:
[tex]1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}[/tex]
1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction above to calculate the enthalpy of combustion (Hcomb) for C8H10
Answer:
[tex]H_{comb}=-4406kJ/mol[/tex]
Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):
[tex]H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}[/tex]
Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:
[tex]H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol[/tex]
Best regards.
A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
Concentration of the weak acid (Ca): 0.187 MpH of the solution: 3.99Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}[/tex]
The acid dissociation constant Ka equals 1.26 × 10–2 for HSO4– (acid 1) and is 5.6 × 10–10for NH4+(acid 2). Predict the net direction of the following reaction: HSO4–(aq) + NH3(aq) SO42–(aq) + NH4+(aq)
Answer:
As K >>> 1, the reaction will shift to the products
Explanation:
To know the direction of any reaction you must calculate the equilibrium constant, K. If K is < 1, the reaction will shift to the reactants and if k > 1 the reaction will shift to the products.
With the reactions:
HSO₄⁻ ⇄ SO₄²⁻ + H⁺ Ka = 1.26x10⁻²
And:
NH₄⁺ ⇄ NH₃⁺ + H⁺ Ka = 5.6x10⁻¹⁰
The inverse reaction:
NH₃⁺ + H⁺ ⇄ NH₄⁺ 1/Ka = 1.8x10⁹
The sum of the reactions:
HSO₄⁻ + NH₃⁺ + H⁺ ⇄ NH₄⁺ + SO₄²⁻ + H⁺ K = 1.26x10⁻² ₓ 1.8x10⁹ = 2.3x10⁷
As K >>> 1, the reaction will shift to the productsWhat is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ksp for Ag₂CO₃ is 8.10 × 10⁻¹²)
Answer:
[tex]\large \boxed{1.64\times 10^{-5}\text{ mol/L }}[/tex]
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
[tex]K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}[/tex]
The amount of the sample in space is referred to as concentration, in this the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].
Concentration Calculation:In chemistry and related sciences, the phrase "concentration" is frequently used. It is a metric for determining how much of one material was mixed with the other.The solution's concentration is indeed the amount of solute absorbed in a given quantity of liquid or solution, following are the calculation of the concentration of [tex]Ag^+[/tex]:Concentration of [tex]Na_2CO_3 = 0.00750 M[/tex]
[tex](CO_3)^{2-} = Na_2CO_3 = 0.00750\ M\\\\Ksp \ \ Ag_2CO_3 =( Ag^{+})^2 (CO_3)^{2-}\\\\8.10 \times 10^{-12} = (Ag^+)^2 \times (0.00750\ M)\\\\(Ag^+)^2 = \frac{(8.10 \times 10^{-12})}{ (0.00750\ M)}\\\\(Ag^+)^2 = 1.08 \times 10^{-9}\ M^2\\\\(Ag^+) = (1.08 \times 10^{-9}\ M^2)^{\frac{1}{2}}\\\\\[(Ag^+)\] = (3.28 \times 10^{-5}\ M)\\\\[/tex]
So, the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].
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A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 33.6 mL of the gas has a mass of 0.087 g. What is the molecular (true) formula for the compound
Answer:
Molecular formula for the gas is: C₄H₁₀
Explanation:
Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g
At STP → 1 atm and 273.15K
1 atm . 0.0336 L = n . 0.082 . 273.15 K
n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)
n = 1.500 × 10⁻³ moles
Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m
Now we propose rules of three:
If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C
58 g of gas (1mol) would have:
(58 g . 0.480) / 0.580 = 48 g of C
(58 g . 0.100) / 0.580 = 10 g of H
48 g of C / 12 g/mol = 4 mol
10 g of H / 1g/mol = 10 moles
The molecular formula of the compound is C4H10.
At STP;
P = 1 atm
T = 273 K
V = 33.6 mL or 0.0336 L
R = 0.082 atmLK-1mol-1
n = ?
Hence;
n = PV/RT
n = 1 atm × 0.0336 L/0.082 atmLK-1mol-1 × 273 K
n = 0.0015 moles
Number of moles = mass/molar mass
Molar mass= Mass/Number of moles
Molar mass = 0.087 g/0.0015 moles
Molar mass = 58 g/mol
Mass of carbon = (58 g × 0.480) / 0.580 = 48 g of C
Mass of hydrogen = (58 g × 0.100) / 0.580 = 10 g of H
Number of moles of carbon = 48 g of C / 12 g/mol = 4 mol
Number of moles of hydrogen = 10 g of H / 1g/mol = 10 moles
Formula of the compound must then be C4H10.
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A student wants to prepare a salt starting with H2SO4. Select all of the compound types that can react with H2SO4 to form a salt.
1. salt
2. acid
3. acid salt
4. basic oxide
5. base
6. metal
7. acidic oxide
Answer:
4 and 6 would work for this
What is the freezing point of a solution of 7.15 g MgCl2 in 100 g of water? K f for water is 1.86°C/m. What is the freezing point of a solution of 7.15 g MgCl2 in 100 g of water? K f for water is 1.86°C/m. -0.140°C -2.80°C -1.40°C -4.18°C
Answer:
THE NEW FREEZING POINT IS -4.196 °C
Explanation:
ΔTf = 1 Kf m
molarity of MgCl2:
Molar mass = (24 + 35.5 *2) g/mol
molar mass = 95 g/mol
7.15 g of MgCl2 in 100 g of water
7.15 g = 100 g
(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3
= 0.715 g /dm3
Molarity in mol/dm3 = molarity in g/dm3 / molar mass
= 0.715 g /dm3 / 95 g/mol
m = 0.00752 mol/ dm3
So therefore:
ΔTf = i Kf m
1 = 3 (1 Mg and 2 Cl)
Kf = 1.86 °C/m
M = 0.752 moles
So we have:
ΔTf = 3 * 1.86 * 0.752
ΔTf = 4.196 °C
The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C