Answer:
The correct answer is 0.4 L.
Explanation:
The mentioned question can be solved by using the equation,
C = K × Pgas--------(i)
Here K is the Henry law constant whose value is 0.158 mol/L/atm, C is the concentration of the gas in liquid state, and Pgas is the partial pressure of the gas.
Now to find the volume of water, the formula to be used is,
PV = nRT-----------(ii)
Here P is the pressure of the gas, V is the volume, R is the universal gas constant whose value is 0.082 Latm/mol/K and T is the temperature.
PgasV = nRT
Pgas = nRT/Vgas
The value of Pgas is inserted in equation (i) we get,
C = K × nRT/Vgas
It is to be noted that C = n/V, here n is the no. of the moles and V is the volume of liquid.
n/Vliquid = K × nRT/Vgas
1/Vliquid = KRT/Vgas
Vliquid = Vgas/KRT--------------(iii)
Based on the given information, the volume of the gas is 1.52 L, the value of K is 0.158 mol/L/atm, the value of R is 0.082 Latm/mol/K and value of T is 21 degree C or 273 + 21 = 294 K.
Now putting the values in equation (iii) we get,
Vliquid = 1.52 L / 0.158 × 0.082 × 294
Vliquid = 1.52 / 3.809
Vliquid = 0.399 or 0.4 L
Hence, the volume of water required to dissolve 1.52 L of gas is 0.4 L.
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]
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
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) If you weigh a hydrated compound before and after heating it. Do you expect the mass to be
more or less than before heating? Explain why.
b) How about if you weigh a bag of popcorn before and after heating it. Does the popped corn
weigh more or less than the unpopped kernels? Explain why.
Answer:
less becuase that moisture adds weight
Explanation:
Phosphoric acid is a triprotic acid. Write the ionization equation(s) and circle any species that are amphoteric.
Answer:
[tex]\mathbf{H_2PO^{-}_4}[/tex] and [tex]\mathbf{HPO^{2-}_4}[/tex] are the amphoteric species.
Explanation:
Phosphorus is a non-metallic element belonging to Group 5 of the Periodic table. As a triprotic acid (i.e an acid that has three dissociable protons that undergo stepwise ionization) . Phosphorus reacts vigorously with hot water to form tetraoxosulphate(V) acid.
i.e
[tex]P_4O_{10(s)} + 6 H_2O_{l} \to4H_3PO_{4(aq)}[/tex]
SO;
As a triprotic acid;
The ionization equations are as follows:
NOTE: Instead of the amphoteric species to be circle as instructed , due to the fact that we are using the LaTex Editor ; we will just need to let them be in bold format which will differentiate them from the rest.
[tex]H_3PO_{4} \leftrightarrow H^+ + H_2PO^{-}_4[/tex]
[tex]\mathbf{H_2PO^{-}_4} \leftrightarrow H^+ + HPO^{2-}_4[/tex]
[tex]\mathbf{HPO^{2-}_4} \leftrightarrow H^+ + PO^{3-}_4[/tex]
Thus; [tex]\mathbf{H_2PO^{-}_4}[/tex] and [tex]\mathbf{HPO^{2-}_4}[/tex] are the amphoteric species.
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]
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!
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:
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.
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.
Will sodium iodide react with bromine to produce sodium bromide and iodine? Why or why not
A.Yes, because bromine has lower activity than iodine
B.Yes, because bromine has higher activity than iodine
C.No, because bromine has lower activity than iodine
D.No,because bromine has higher activity than iodine
Answer:
C.No, because bromine has lower activity than iodine
Explanation:
Hello,
In this case, for answering the given question we should remember that the higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, we can say that the described reaction is not possible due to the fact that bromine is less active. For that reason answer is C.No, because bromine has lower activity than iodine.
Best regards.
Answer:
Yes, because bromine has lower activity than iodine
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
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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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.
1. Arrange the following compounds in order from highest to lowest boiling point.
hexanol (C6H14O), dichloromethane (CH2Cl2), CCl4, butene (C4H8), propane (C3H8)
2. Identify the three true statements:
a. Hydrocarbons exhibit only dispersion forces.
b. Stronger intermolecular forces usually correlate with higher boiling points.
c. Hydrogen bonds require carbon, hydrogen, and a halogen.
d. Dipole-dipole interactions are stronger than dispersion forces and hydrogen bonds.
e. Boiling point generally increases with total number of electrons in molecules due to increased strength of dispersion forces.
Answer:
1. Hexanol (C6H14O) > Carbon tetrachloride (CCl4) > Dichloromethane (CH2Cl2) > butene (C4H8) > propane (C3H8)
2. a. b and e
Explanation:
1. Arrangement of given compounds in order from highest to lowest boiling point are as following:
hexanol (C6H14O): 157 degree celcius
carbon tetrachloride (CCl4): 76.72 degree celcius
dichloromethane (CH2Cl2): 40 degree celcius
butene (C4H8): -6.3 degree celcius
propane (C3H8): -42 degree celcius
Hexanol (C6H14O) > Carbon tetrachloride (CCl4) > Dichloromethane (CH2Cl2) > butene (C4H8) > propane (C3H8).
2. True statements are:
a. Hydrocarbons have only dispersion forces because hydrocarbons are very non-polar and are highly symmetric molecules.
b. Stronger intermolecular forces require higher boiling points because the bond between the molecules is very strong and require high energy and boiling point to break that strong bond.
e. Because the size of molecules increases with increase in number of electrons and dispersion force increases the attraction of the molecules to each other. Hence require high boiling point to separate them from each other.
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
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
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 productsDeuterium is an isotope of hydrogen (H) that has:
Answer:
D2 has 1 neutron
and 1 proton
A solution of 49.0% H2SO4 by mass has a density of 1.39 g cm−3 at 293 K. A 22.6 cm3 sample of this solution is mixed with enough water to increase the volume of the solution to 88.5 cm3 .
Find the molarity of sulfuric acid in this solution.
Answer:
The molarity of the sulfuric acid in the solution is 1.77 M.
Explanation:
The molarity of the sulfuric acid in the solution can be found using the following equation:
[tex] C_{i}V_{i} = C_{f}V_{f} \rightarrow C_{f} = \frac{C_{i}V_{i}}{V_{f}} [/tex]
Where:
[tex]C_{i}[/tex]: is the initial concentration of the acid
[tex]V_{i}[/tex]: is the initial volume of the solution = 22.6 cm³
[tex]V_{f}[/tex]: is the final volume of the solution = 88.5 cm³
The initial concentration of the H₂SO₄ is:
[tex] C_{i} = \frac{n}{V} = \frac{m}{M*V} = \frac{d*\% ^{m}_{m}}{M} [/tex]
Where:
n: is the number of moles
m: is the mass
M: is the molar mass = 98.079 g/mol
d: is the density of the acid = 1.39 g/cm³
%: is the percent by mass = 49.0 %
[tex] C_{i} = \frac{1.39 \frac{g}{cm^{3}}*\frac{1000 cm^{3}}{1 L}*\frac{49 g}{100 g}}{98.079 \frac{g}{mol}} = 6.94 M [/tex]
Finally, the final concentration of H₂SO₄ after the dilution is:
[tex] C_{f} = \frac{6.94 M*22.6 cm^{3}}{88.5 cm^{3}} = 1.77 M [/tex]
Therefore, the molarity of the sulfuric acid in the solution is 1.77 M.
I hope it helps you!
A young scientist wants to focus on drug creation. What is one of the first
things he should do?
The given question is incomplete as it lacks the appropriate options, however, the options related to this question is as follows:
a)Test a composite material in the lab to see how it can be improved.
b)Look into trends of various diseases and their effects on society.
c)Research available drugs for treating illnesses such as cancer and heart disease, as well as the side effects of these drugs.
d)Make a plan for how to conquer a disease with drugs.
Answer:
The correct answer is - b)Look into trends of various diseases and their effects on society.
Explanation:
The young scientist who wants to focus on creating drugs to treat disease must think and look for the diseases that are needed to be treated at the moment or affecting the population most and how it affects the society and then plan how to do it.
So, the scientist needs to check the trends of various disease and their effect on society for developing understanding which disease needs to be cure first and how it affects.
Thus, the most appropriate answer is - option B.
Answer:
b)Look into trends of various diseases and their effects on society.
Explanation:
ap3x
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.
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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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.
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!
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
Calculate the root-mean-square velocity for a hydrogen molecule at
-15.0 °C. The universal gas constant, R=8.314 J/mol.K.
Answer:
56.5 m/s
Explanation:
Step 1: Convert the temperature to Kelvin
We will use the following expression.
K = °C + 273.15
K = -15.0°C + 273.15 = 258.2 K
Step 2: Calculate the root-mean-square velocity
The root-mean-square velocity is the square root of the average square velocity. We can calculate it using the following expression.
[tex]v_{rms} = \sqrt{\frac{3RT}{M} }[/tex]
where,
R: ideal gas constantT: absolute temperatureM: molar mass of the gas[tex]v_{rms} = \sqrt{\frac{3 \times \frac{8.314J}{mol.K} \times 258.2K }{2.02g/mol} } = 56.5m/s[/tex]
The root-mean-square velocity for a hydrogen molecule at the given temperature is 1,793.74 m/s.
The given parameters;
temperature of the hydrogen, t = -15 ⁰Cmolecular mass of hydrogen, m = 2 g/moluniversal gas constant, R = 8.314 J/mol.KThe root-mean-square velocity for a hydrogen molecule is calculated as follows;
[tex]v_{rms} = \sqrt{\frac{3RT}{m} } \\\\[/tex]
where;
T is the temperature of the hydrogen = -15 + 273 = 258 Km is the mass of the hydrogen = 2 x 10⁻³ kg/mol[tex]v_{rms} = \sqrt{\frac{3 \times 8.314 \times (-15+273)}{2\times 10^{-3}} } \\\\v_{rms} = 1,793.74 \ m/s[/tex]
Thus, the root-mean-square velocity for a hydrogen molecule at the given temperature is 1,793.74 m/s.
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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]
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.