Answer:
296.1g of Ag is the maximum amount of silver
Explanation:
A solution of 6.3% Ag by mass contains 6.3g of Ag per 100g of solution. Thus, you need to calculate the mass of the solution and then, the mass of Ag present in solution, thus:
Mass of solution:
Assuming a density of 1g/mL:
[tex]4.7L \frac{1000mL}{1L} \frac{1g}{mL} = 4700g[/tex]
If the solution contains 6.3g of Ag per 100g of solution, the mass of Ag in 4700L is:
4700L × (6.3g Ag / 100g) =
296.1g of Ag is the maximum amount of silverg A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the radiation wavelength is 12.5 cm, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC
Answer:
The total photons required = 5.19 × 10²⁸ photons
Explanation:
Given that:
the radiation wavelength λ= 12.5 cm = 0.125 m
Volume of the container = 0.250 L = 250 mL
The density of water = 1 g/mL
Density = mass /volume
Mass = Volume × Density
Thus; the mass of the water = 250 mL × 1 g/mL
the mass of the water = 250 g
the specific heat of water s = 4.18 J/g° C
the initial temperature [tex]T_1[/tex] = 20.0° C
the final temperature [tex]T_2[/tex] = 99° C
Change in temperature [tex]\Delta T[/tex] = (99-20)° C = 79 ° C
The heat q absorbed during the process = ms [tex]\Delta T[/tex]
The heat q absorbed during the process = 250 g × 4.18 J/g° C × 79° C
The heat q absorbed during the process = 82555 J
The energy of a photon can be represented by the equation :
= hc/λ
where;
h = planck's constant = [tex]6.626 \times 10^{-34} \ J.s[/tex]
c = velocity of light = [tex]3.0 \times 10^8 \ m/s[/tex]
= [tex]\dfrac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{0.125}[/tex]
= [tex]1.59024 \times 10^{-24}[/tex] J
The total photons required = Total heat energy/ Energy of a photon
The total photons required = [tex]\dfrac{82555 J}{1.59024 \times 10^{-24}J}[/tex]
The total photons required = 5.19 × 10²⁸ photons
A volumetric flask contains 25.0 mL of a 14% m/V sugar solution. If 2.5 mL of this solution is added to 22.5 mL of distilled water, what is the % m/V of the new solution.
Answer:
The new solution is 1.4% m/V
Explanation:
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:
[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]
We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
You can learn more about dilution here: https://brainly.com/question/13844449
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/
Answer:
10.328 m
Explanation:
normal atmospheric pressure = 101325 Pa
density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3
pressure = pgh
where p = density
g = acceleration due to gravity = 9.81 m/s^2
h = height of column
imputing values, we have
101325 = 1000 x 9.81 x h
height of column h = 101325/9810 = 10.328 m
Recall that when a reaction is at equilibrium, the forward and reverse reactions occur at the same rate. To illustrate this idea, consider the reaction of A (small, red spheres) and B (large, blue spheres) to form AB.
A+B ⇌ AB
Notice that the reaction never stops. Even after several minutes, there is A and B left unreacted, and the forward and reverse reactions continue to occur. Also note that amounts of each species (i.e., their concentrations) stay the same.
Required:
What is the value of the equilibrium constant for this reaction?
Assume each atom or molecule represents a 1 M concentration of that substance.
Answer:
Equilibrium constant Kc = [x]² / [A - x] [B - x]
Explanation:
The equilibrium constant is defined as the ratio of the concentration of the products to that of the reactants at equilibrium
ie Kc = [products] / [reactants].
The balanced equation of the reaction is given as : A + B ⇄ AB
At the beginning of the reaction,
Initial concentration I = A = 1M
B = 1M
AB = 0M
After a period of time and assuming 'x' to be the concentration of product AB formed, the concentrations become
C = reactant A = [A - x] M
rectant B = [B - x] M
Product AB = [x] [x] M
At equilibrium, the concentrations are,
E = rectant A = [A - x] M
reactant B = [B - x] M
product AB = [x]² M
therefore , the equilibrium constant, Kc = [products]/[reactants]
= [x]² / [A - x] [B - x]
Based on the Valence Shell Electron Pair Repulsion Theory (or VSEPR), molecules will arrange to keep the following as far apart as physically possible
a) mobile electrons
b) valence electron pairs
c) inner shell electrons
d) the electrons closest to the nuclei
Answer:
B. Valence Electron Pairs
Explanation:
Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.
Consider the following precipitation reaction occurring in aqueous solution:
3 SrCl2(aq)+2 Li3PO4(aq) →Sr3(PO4)2(s)+6 LiCl(aq)
Write the complete ionic equation and the net ionic equation for this reaction.
Answer:
[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]
First equation is the complete ionic equation.
Second equation is the net ionic equation.
7. An element's most stable ion forms an ionic compound with chlorine having the formula XCl2. If the ion of element X has a mass of 89 and 36 electrons, what is the identity of the element, and how many neutrons does it have
Answer:
The element is strontium and the number of neutrons it have is 51.
Explanation:
Based on the given information, the ionic compound is,
XCl₂ ⇔ X₂⁺ + 2Cl⁻
X2+ is the ion of the mentioned element
As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,
No. of electrons = atomic number - charge
36 = atomic number - 2
Atomic number = 38
Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.
Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,
No. of neutrons = mass number - atomic number
= 89 - 38
= 51 neutrons.
The Ka1 value for oxalic acid is 5.9 x10-2 , and the Ka2 value is 4.6 x 10-5 . What are the values of Kb1 and Kb2 of the oxalate ion
Answer:
2.17x10⁻¹⁰ = Kb1
1.69x10⁻¹³ = Kb2
Explanation:
Oxalic acid, C₂O₄H₂, has two intercambiable protons, its equilibriums are:
C₂O₄H₂ ⇄ C₂O₄H⁻ + H⁺ Ka1 = 5.9x10⁻²
C₂O₄H⁻ ⇄ C₂O₄²⁻ + H⁺ Ka2 = 4.6x10⁻⁵
Oxalate ion, C₂O₄²⁻, has as equilibriums:
C₂O₄²⁻ + H₂O ⇄ C₂O₄H⁻ + OH⁻ Kb1
C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2
Also, you can know: KaₓKb = Kw
Where Kw is 1x10⁻¹⁴
Thus:
Kw = Kb2ₓKa1
1x10⁻¹⁴ =Kb2ₓ4.6x10⁻⁵
2.17x10⁻¹⁰ = Kb1And:
Kw = Kb1ₓKa2
1x10⁻¹⁴ =Kb1ₓ5.9x10⁻²
1.69x10⁻¹³ = Kb1
That is because the inverse reaction of, for example, Ka1:
C₂O₄H⁻ + H⁺ ⇄ C₂O₄H₂ K = 1 / Ka1
+ H₂O ⇄ H⁺ + OH⁻ K = Kw = 1x10⁻¹⁴
=
C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2 = Kw × 1/Ka1
what is the polarity of black pepper
Answer:
Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.
Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.
It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.
How are scientific questions answered?
A. Through observing and measuring the physical world
B. Through testing a theory about the physical world
c. Through forming a hypothesis about the question
D. Through predicting a solution about the question
SUBM
Answer:
Option B
Explanation:
Scientific question are answered through experimentation, through testing the theory about the physical world.
Answer: its A
through observing and measuring the physical world
Explanation:
If an electron has a principal quantum number (n) of 7 and an angular momentum quantum number (l) of 1, the subshell designation is ________
Answer:
7p
Explanation:
principal quantum number is 7
n=7( principle shell)
angular momentum quantum number gives sub shell
l = 1 means it is p orbital
so answer is 7p orbital
How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water
Answer:
WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.
Explanation:
Equation for the reaction:
K2SO4 + H20 ------->2 K+ + SO4^2-
When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.
1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-
At STP, 1 mole of K2SO4 will be the molar mass of the substance
Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol
Molar mass = 174 g/mol
So therefore;
1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion
When 63.7 g is used; we have:
174 g = 2 moles of K+
63.7 g = ( 63.7 * 2 / 174) moles of K+
= 0.73 moles of K+
Forr sulfate ion, we have:
174 g = 1 mole ofSO4^2-
63.7 g = (63.7 * 1 / 174) moles of SO4^2-
= 0.366 moles of SO4^2-
In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.
The NaOH solution is standardized (or its true concentration) is found by reacting it with KHSO4. One of the two products from when NaOH reacts with KHSO4 is H2O. The other product is is a salt consisting of what?
a. NaK (aq)
b. (aq)
c. NaS (aq)
d. None of the above
What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12
Answer:
pH = 3.39
Explanation:
The equilibrium in water of ascorbic acid (With its conjugate base) is:
H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)
Where the acidic dissociation constant is written as:
Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]
H₂O is not taken in the Ka expression because is a pure liquid.
As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:
[H₂C₆H₆O₆] = 2.5x10⁻³M - X
[HC₆H₆O₆⁻] = X
[H₃O⁺] = X
Replacing in the Ka expression:
7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]
1.975x10⁻⁷ - 7.9x10⁻⁵X = X²
0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷
Solving for X:
X = -0.00048566→ False solution, there is no negative concentrations
X = 0.00040666 → Right solution
As [H₃O⁺] = X, [H₃O⁺] = 0.00040666
pH is defined as -log [H₃O⁺];
pH = -log 0.00040666,
pH = 3.39Therapeutic drugs generally need to have some hydrophobic and hydrophilic components to be able to effectively reach their target organs and tissues given there are aqueous and nonaqueous parts of the body. The degree to which a compound is hydrophobic and hydrophilic can be determined by measuring its relative solubility in water and octanol, C8H17OH, and water. To do this, a sample of the compound is added to a mixture of water and octanol and mixed well. Water and octanol are immiscible so after the mixture settles, the concentration of the compound in water and the concentration of the compound in octanol is measured. The ratio of the concentrations is called the partition ratio:
The question is incomplete as some part is missing:
concentration in octanol Partition Ratio = concentration in water
a) What are the intermolecular forces of attraction between octanol molecules? Explain.
b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.
c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.
d) Would nonane (figure 2) be more soluble in water or octanol? Explain.
e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.
f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.
Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C
Answer:
1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.
2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).
3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.
4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.
5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.
What is the systematic name for the given compound? (R)-2-bromobutane (S)-2-bromobutane (R)-2-butyl bromide (S)-2-butyl bromide
Answer:
See explanation below
Explanation:
IUPAC came up with the idea of an unambiguous system of nomenclature for organic compounds. This unambiguous system relates the structure of a compound with its name. Thus, IUPAC has established a worldwide standard for the unambiguous naming of organic compounds. Scientists all over the world can now have a uniform system of nomenclature for compounds in order to facilitate easy communication of scientific information.
The systematic names of the following compounds listed in the question are shown below;
(R)-2- butyl bromide has the systematic name (R)-2-bromobutane
(S)-2-butyl bromide has the systematic name (S)-2-bromobutane
This unified system of nomenclature avoids the confusion created by the use of different trivial names in deferent localities and by various scientific academies. This is a major advantage of the systematic nomenclature.
Write the following isotope in nuclide notation: oxygen-14
Answer:
[tex]14\\8[/tex]O
Explanation:
The top number always represents the mass number.
The bottom number always represents the atomic number.
The element always goes after the numbers.
If charge is present, that comes after the element.
Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng
Answer:
143pm is the radius of an Al atom
Explanation:
In a face centered cubic structure, FCC, there are 4 atoms per unit cell.
First, you need to obtain the mass of an unit cell using molar mass of Aluminium and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.
Mass of an unit cell
As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:
4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g
Edge length
As density of aluminium is 2.71g/cm³, the volume of an unit cell is:
1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³
And the length of an edge of the cell is:
∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m
Radius:
As in FCC structure, Edge = √8 R, radius of an atom of Al is:
4.044x10⁻¹⁰m = √8 R
1.430x10⁻¹⁰m = R.
In pm:
1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =
143pm is the radius of an Al atomThe radius of the atom of Al in the FCC structure has been 143 pm.
The FCC lattice has been contributed with atoms at the edge of the cubic structure.
The FCC has consisted of 4 atoms in a lattice.
The mass of the unit cell of Al can be calculated as:[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole
4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles
The mass of 1 mole Al has been 26.98 g/mol.
The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g
The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.
The volume of the Al cell can be calculated as:Density = [tex]\rm \dfrac{mass}{volume}[/tex]
Volume = Density × Mass
The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g
The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]
The volume of the cube has been given as:Volume = [tex]\rm edge\;length^3[/tex]
6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]
Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm
Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm
Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
In an FCC lattice structure, the radius of the atom can be given by:Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
1 m = [tex]\rm 10^1^2[/tex] pm
1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.
The radius of the atom of Al in the FCC structure has been 143 pm.
For more information about the FCC structure, refer to the link:
https://brainly.com/question/14934549
help please !!!!!!!!
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
Question 23
1 pts
When solutions of AgNO3 and NaOH react, the balanced molecular equation is:
2 AgNO3(aq) + 2NaOH(aq) O--> Ag2O(s) + 2 NaNO3(aq) + H20(1)
How much Ag2O is produced when 0.200 g of AgNO3 and 0.200 g of NaOH react?
a. 0.127 g
c. 0.273 g
b. 0.136 g
d. 0.400 g
OB
OC
OA
OD
Answer:
Option B. 0.136 g
Explanation:
The balanced equation for the reaction is given below:
2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)
Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:
Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol
Mass of AgNO3 from the balanced equation = 2 x 170 = 340g
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the balanced equation = 2 x 40 = 80g
Molar mass of Ag2O = (108x2) + 16 = 232g/mol
Mass of Ag2O from the balanced equation = 1 x 232 = 232g
Summary:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH.
Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.
From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.
Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.
In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.
The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted to produce 232g of Ag2O.
Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.
Therefore, 0.136g of Ag2O was produced from the reaction.
Calculate the work (kJ) done during a reaction in which the internal volume expands from 28 L to 51 L against an outside pressure of 4.9 atm.
Answer:
W= -11KJ
Explanation:
Given:
volume expands from 28 L to 51 L
pressure =4.9 atm.
We will need to Convert the pressure to Pascal SI
But 1 atm = 101,325 Pa.
Then,
Pressure= (4.9*101323)/1atm = 5*10^5 pa
Then we need to Convert the volumes to cubic meters
But we know that1 m³ = 1,000 L.
V1= 28L * 1m^3/1000L = 0.028m^3
V2=51L × 1m^3 /1000L =0.051m^3
The work done during the expansion of a gas can be calculated as
W= -P(V2-V1)
W= - 5*10^5(0.051m^3 - 0.028m^3)
W= -1.1× 10^4J
Then we can Convert the work to kiloJoule
But1 kJ = 1,000 J.
W= -1.1× 10^4J× 1kj/1000J
= -11KJ
Calcium carbide, CaC2, reacts with water to form calcium hydroxide and the flammable gas ethyne (acetylene) in the reaction: What mass of ethyne can be produced
Answer:
1 mole of CaC₂ will produce 26g of C₂H₂ or 64.1g of CaC₂ will produce 26g of C₂H₂
Explanation:
Hello,
To solve this question, we'll require a balanced chemical equation of reaction between calcium carbide and water.
Equation of reaction
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
Molar mass of calcium carbide (CaC₂) = 64.1g/mol
Molar mass of water (H₂O) = 18g/mol
Molar mass of calcium hydroxide (Ca(OH)₂) = 74g/mol
Molar mass of ethyne (C₂H₂) = 26g/mol
From the equation of reaction, 1 mole of CaC₂ will produce 1 mole of C₂H₂
1 mole of CaC₂ = mass / molar mass
Mass = 1 × 64.1
Mass = 64.1g
1 mole of C₂H₂ = mass / molar mass
Mass = 1 × 26
Mass = 26g
Therefore, 1 mole of CaC₂ will produce 26g of C₂H₂
Note: this is a hypothetical calculation since we were not given the initial mass of CaC₂ that starts the reaction
Which one of the following would have the largest dispersion forces?
A) CH3CH2SH
B) CH3NH2
C) CH4
D) CH3CH3
Answer:
A) CH3CH2SH
Explanation:
Dispersion forces are weak attractions found between non-polar and polar molecules. The attractions here can be attributed to the fact that a non-polar molecule sometimes become polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant. If this happens, the molecule has a temporary dipole. This dipole can induce the neighbouring molecules to be distorted and form dipoles as well. The attractions between these dipoles constitute the Dispersion Forces.
Therefore; the greater the molar mass of a compound or molecule, the higher the Dispersion Force. This implies that the compound or molecule with the highest molar mass have the largest dispersion forces.
Now; for option (A)
CH3CH2SH
The molar mass is :
= (12 + (1 × 3 ) +12 + (1 ×2) + 32+1)
= (12 + 3+ 12 + 2 + 32 + 1)
= 62 g/mol
For option (B)
CH3NH2
The molar mass is:
= (12 + (1 × 3 ) +14 + (1 × 2)
= (12 + 3 + 14 + 2)
= 31 g/mol
For option (C)
CH4
The molar mass is :
= 12 + (1 × 4)
= 12 + 4
= 16 g/mol
For option (D)
CH3CH3
The molar mass is :
= 12 + ( 1 × 3 ) + 12 + ( 1 × 3)
= 12 + 3 + 12 + 3
= 30 g/mol
Thus ; option (A) has the highest molar mass, as such the largest dispersion force is A) CH3CH2SH
How many water molecules are in a block of ice containing 0.500 mol of water (H2O)?
Answer:
3.01 × 10²³ molecules
Explanation:
Step 1: Given data
Moles of water (n): 0.500 mol
Step 2: Calculate the molecules of water present in 0.500 moles of water
In order to perform this calculation, we will use the Avogadro's number: in 1 mole of water there are 6.02 × 10²³ molecules of water.
0.500 mol × (6.02 × 10²³ molecules/1 mol) = 3.01 × 10²³ molecules
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]
Why does the excess of base used in these eliminations favor the E2 over the E1 mechanism for elimination
Answer:
The base is involved in the rate determining step of an E2 reaction mechanism
Explanation:
Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;
Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.
For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;
Rate = k[alkyl halide] [base]
The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.
Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.
Determine the rate of a reaction that follows the rate law:
rate = k[A]”[B]", where:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Answer:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Explanation:
rate = k[A]”[B]"
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
What is meant by rate of a reaction ?Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.
Here,
The concentration of A, [A] = 1 M
The concentration of B, [B] = 3 M
The partial order with respect to A, m = 2
The partial order with respect to B, n = 1
The rate constant of the reaction, k = 1.5
The rate of the reaction,
r = k[A]^m [B}^n
r = 1.5 x 1² x 3
r = 4.5 mol L⁻¹s⁻¹
Hence,
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
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Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light?
1. Gamma rays
2. Infrared radiation
3. Ultraviolet liht
4. X-rays
5. Microwaves
6. Radio waves
Answer:
3,4,1 and 6,5,2
Explanation:
In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;
Radio waves
Microwaves
Infrared waves
Visible light rays
Ultraviolet rays
X-rays
Gamma rays
(a simple mnemonic is RMIVUXG)
A substance used as a cleaner and a fuel is 37.48% C, 49.93% O and 12.58% H by mass. A 0.2804-g sample of the substance occupies a volume of 250.0 mL when it is vaporized at 75o C and 1.00 atm of pressure.
R = 0.0821 L atm/ mol K
a) This compound can be made by combining gaseous carbon monoxide with hydrogen gas (with this compound as the only product). What is the maximum mass of this compound that can be prepared if 8.0 kg of hydrogen gas react with 59.0 kg of carbon monoxide gas?
b) If 59.6 kg of the product is actually produced, given the reaction described in (a), what is the percent yield?
c) This compound (the substance you identified in part a) is a potential replacement for gasoline. The products of the complete combustion of this fuel are the same as those for the complete combustion of a hydrocarbon (CO2 and H2O). Calculate the volume of CO2 produced at 27o C and 766 mmHg when 1.00 gallon of this fuel is completely combusted. The density of the fuel is 0.7914 g/mL. 1 gallon = 3.785 liters
d) A claim was made that this fuel is better for the environment because it produces less CO2 per gallon than gasoline, which can be represented by the formula C8H18 (octane). Is this claim true? Octane has a density of 0.6986 g/mL
Answer:
Explanation:
We shall find out the molecular formula of the substance .
Ration of number of atoms of C , O and H
= [tex]\frac{37.48}{12} :\frac{49.93}{16} :\frac{12.58}{1}[/tex]
= 3.12 : 3.12 : 12.58
= 1 : 1 : 4
volume of gas at NTP
= 250 x 273 / 350 mL .
= 195 mL .
Molecular weight of the substance = .2804 x 22400 / 195 g
= 32. approx
Let the molecular formula be
(COH₄)n
n x 32 = 32
n = 1
Molecular formula = COH₄
The compound appears to be CH₃OH
a )
CO + 2H₂ = CH₃OH
28g 4g 32g
59 8
For 8 kg hydrogen , CO required = 56 kg
CO is in excess . hydrogen is the limiting reagent .
mass of product formed
= 32 x 8 / 4
= 64 kg
b )
percentage yield = product actually formed / product to be formed theoretically x 100
= 59.6 x 100 / 64
= 93.12 %
c )
2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .
64 g 2 x 22.4 L
Gram of gas in 1 gallon of fuel
= .7914 x 3785
= 2995.5 g
CO₂ produced at NTP by 2995.5 g CH₃OH
= 2 x 22.4 x 2995.5 / 64 L
= 2096.85 L
At 27° C and 766 mm Hg , this volume is equal to
2096.85 x 300 x 760 / 273 x 766
= 2286.18 L .
d )
C₈H₁₈ = 8CO₂
114g 8 x 22.4 L
gram of fuel per unit gallon
= .6986 x 3785
= 2644.2g
gram of CO₂ produced by 1 gallon of fuel at NTP
= 8 x 22.4 x 2644.2 / 114
= 4156.5 L
So it produces more CO₂ .
Identify the correctly written chemical reaction
A. Reactant + Reactant = Product
B. Reactant + Reactant → Product + Product
C. Reactant + Product → Reactant + Product
D. Product + Product Reactant + Reactant
Answer:
B. Reactant + Reactant -> Product + Product
Explanation:
Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction
Products are substances that are produced as a result of the reaction
Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving. In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards
And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.
To keep things simple:
Products can't be at the beginning of a reaction since they weren't formed yet.
Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products
As such, only one possible chemical reaction would follow that reasoning:
Reactant + Reactant -> Product + Product
Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.
What is a chemical equation?A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.
In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.
Hence, option B is correct.
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