Answer:
92.75%
Explanation:
The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:
[tex]\mathtt{2Al + 2KOH + 4H_2SO_4 +2H_2O \to 2KAl(SO_4)_2 2H_2O +3H_2}[/tex]
From above; we will see that 2 moles of Aluminium react with sulphuric acid and water to produce 2 moles o aluminium alum.
Therefore, the theoretical yield can be determined as:
[tex]=0.4395 \ g Al \times \dfrac{1 \ mol \ Al}{27 \ g Al }\times \dfrac{2 \ mol \ KAl(SO_4)_2}{2 \ mol \ Al}\times \dfrac{294.23 \ KAl(SO_4)_2}{1 \ mol \ KAl(SO_4)_2}[/tex]
= 4.789g of [tex]KAl(SO_4)_2[/tex]
To find the percent yield, we need to divide the actual yield by the theoretical yield and then multiply it with 100.
∴
percent yield = ( mass of alum(g)/theoretical yield(g) ) × 100
percent yield = ( 4.789g / 5.1629g ) × 100%
percent yield = 0.9275 × 100%
percent yield = 92.75%
Thus, the percent yield of the experiment 92.75%
Which of the following best describes the structure of a nucleic acid?
a. Carbon ring(s)
b. Globular or fibrous
c. Single or double helix
d. Hydrocarbon(s)
Which of the following elements has the largest atomic radius? Bromine, Barium, Magnsium, Zinc
Answer:
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
Explanation:
A geochemist has determined by measurements that there are moles of tin in a sample of cassiterite. How many moles of oxygen are in the sample
The question is incomplete, the complete question is:
This is the chemical formula for cassiterite (tin ore): SnO2 A geochemist has determined by measurements that there are 13. moles of tin in a sample of cassiterite. How many moles of oxygen are in the sample? Round your answer to 2 significant digits.
Answer:
Explanation:
In SnO2, there are two moles of oxygen for each mole of tin.
Hence, if there are 13 moles of tin, then we should have 13 * 2 moles of oxygen. This gives us 26 moles of oxygen.
Hence there are 26 moles of oxygen.
Three colorless solutions in test tubes, with no labels, are in a test tube rack on the laboratory bench. Lying beside the tests tubes are three labels : 0.10 M Na2CO3, 0.10 M HCL, and 0.10 M KOH. You are to place the labels on the test tubes using only the three solutions present. Here are your tests:
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm.
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Identify the labels for test tubes 1, 2, and 3
Answer:
Test tube 1 0.10 M HCL
Test tube 2 0.10 M KOH
Test tube 3 0.10 M Na2CO3
Explanation:
From the question we are told that
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm
Generally this warmth is as a result of a reaction between an acid and a base and the acid is 0.10 M HCL and the base is 0.10 M KOH , the heat generated is know as the heat of neutralization,
The reaction is
[tex]HCl_{(aq)} + KOH_{(aq)} \rightarrow KCl_{(aq)} + H_2O_{(l)} + \Delta H[/tex]
We are also told from the reaction that
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Generally carbon dioxide gas is produced is as a result of a reaction between the acid HCl and Na2CO3.
The reaction is
[tex]2HCl -{(aq)} + Na_2 CO_3_{(aq)} \rightarrow 2 NaCl _{(aq)} + CO_2_{(g)} + H_2O_{(l)}[/tex]
Hence from this explanation above we see that the solution in test tube 1 is 0.10 M HCL while solution in test tube 2 is 0.10 M KOH and then solution in test tube three is 0.10 M Na2CO3
Which element will gain three electrons to form an anion?
A. aluminum
B. chromium
C. iron
D. nitrogen
Answer:
D represents the element nitrogen which will gain three valence electrons forming a 3 ion.
Answer:
answer is
Explanation:
D
Gases A and B are confined to a cylinder and piston and react to form product C. As the reaction occurs, the system loses 1189 J of heat to surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 311 J of work on the system. What is the change in the internal energy of the system
Answer:
The change in the internal energy of the system -878 J
Explanation:
Given;
energy lost by the system due to heat, Q = -1189 J (negative because energy was lost by the system)
Work done on the system, W = -311 J (negative because work was done on the system)
change in internal energy of the system, Δ U = ?
First law of thermodynamics states that the change in internal energy of a system (ΔU) equals the net heat transfer into the system (Q) minus the net work done by the system (W).
ΔU = Q - W
ΔU = -1189 - (-311)
ΔU = -1189 + 311
ΔU = -878 J
Therefore, the change in the internal energy of the system -878 J
If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction
Answer:
7.71 × 10⁻⁴ M/s
Explanation:
The initial rate of the reaction can be expressed by using the formula:
[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]
where;
Pressue P = 1.00 atm
Volume V =5.74mL = (5.74 /1000) L
Rate R = 0.082 L atm/mol.K
Temperature = 298 K
[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]
= 2.35 × 10⁻⁴ mol
Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]
Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]
Δ[O₂] = 0.04626 M
The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
= [tex]\dfrac{0.04626}{60}[/tex]
= 7.71 × 10⁻⁴ M/s
Which of the items below is a colloid?
a.fruit salad b.gelatin c.lacquer
Answer:
b. gelatin
Explanation:
a homogeneous noncrystalline consisting of large molecules or ultramicroscopic particles of one substance.
2) (3 pts) Convert 85 oF to oC.
Answer:
The answer will be 29.4˚C.
Explanation:
Using the formula 5(˚F-32)/9, plug in the numbers and you'll get 29.4˚C.
what kind of bonds are there in H2O?
Answer: it is covalent and there are 2 hydrogen molecules and 1 oxygen molecule.
Explanation: it just is
What does reflection mean?
Answer:
the throwing back by a body or surface of light, heat, or sound without absorbing it.
"the reflection of light"
Explanation:
i hope this helps
In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.
Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?
Answer:
a) - 0.2 M
b) - 0.2 M
c)- 0
Explanation:
The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:
MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol
a). Molarity = moles CuBr₂/1 L solution
moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol
Volume in L = 375 mL x 1 L/1000 mL = 0.375 L
M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M
b). When is added to water, CuBr₂ dissociates into ions as follows:
CuBr₂ ⇒ Cu²⁺ + 2 Br⁻
We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:
0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M
c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).
The wood in my house is crumbling. *
Problem
Hypothesis
Law
Theory
Answer:
Problem
Explanation:
The given statement is a problem. It states the problem that the house of the speaker has been undergoing with. This problem gives rise to the Hypothesis in which the 'why' question is asked. The reason of the crumbling of the wood is stated in the hypothesis. Any reason placed of the happening of the event is stated to be hypothesis.
Rahul and Manav each were given a mixture of iron filings and sulphur powder. Rahul heated the mixture strongly and a new substance was formed. Write three points of difference between the two.Required to answer.
Answer:
sure
Explanation:
The substance formed after heating the mixture of that of Rahul is caleed a compound. Whereas, Manav's mixture still remains in its current stae that is a heterogeneous mixture.
The compound formed is in black in color whereas the mixture is a mix of brownish-red and yellow.
The compound is a homogeneous mixture whereas the mixture is a heterogenous mixture because of its uneven distribution.
How long would it take a bus traveling 52 km/h to travel 130 km
Answer:
2 and a half hours
Explanation:
Time is equal to distance over speed
A piece of metal has a mass of 0.650 kilograms, has a width of 0.136 meters, and has a length of 0.0451 meters.Part A: If the metal’s volume is 291 cm3, what is the height of the metal in centimeters? (The width & length values given above are in a different unit!)
Part B: What is the density of this piece of metal?
Answer:
height = 4.74 cm
density = 2.23 g/ cm³
Explanation:
Mass of metal = 0.650 kg (650 g)
Width = 0.136 m
Length = 0.0451 m
Volume of metal = 291 cm³
Height in cm = ?
density of metal =?
Solution:
Width = 0.136 m (0.136 m×100 cm/1m = 13.6 cm)
Length = 0.0451 m (0.0451 m×100 cm/1m = 4.51 cm)
First of all we will calculate the height:
Volume = height× width× length
291 cm³ = h × 13.6 cm × 4.51 cm
291 cm³ = h × 61.34 cm²
h = 291 cm³ / 61.34 cm²
h = 4.74 cm
Density:
d = m/v
d = 650 g/291 cm³
d = 2.23 g/ cm³
9 What 11 letter word describes the chemical balance within an organism?
___________________________________
10 What 7 letter word describes the waterproof outer layer for soft-bodies creatures?
___________________________________
11 What word starting with the letter "C" is the liquid material found within cells?
___________________________________
12 What word starting with the letter "C" is a process of organizing things into groups scientifically?
___________________________________
Answer:
9.) Homeostasis
10.) Epidermis
11.) Cytoplasim
12.) Classification
For the combustion of methane presented in Example 5.4, the chemical reaction is CH4 +2O2 →CO2 +2H2O Suppose that methane flows into a burner at 30 gmol/s, while oxygen flows into the same burner at 75 gmol/s. If all the meth- ane is burned and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? Hint 1: Does the fact that all the methane is burned mean that all the oxygen is burned also? Hint 2: Find the molar flow rate of each component gas in the outlet gas ("flue gas").
Answer:
[tex]x_{CO_2}^{out} =0.25[/tex]
Explanation:
Hello.
In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:
[tex]CH_4:f_{CH_4}^{out}=f_{CH_4}^{in}-\epsilon \\\\O_2:f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon\\\\CO_2:f_{CO_2}^{out}=\epsilon\\\\H_2O:f_{H_2O}^{out}=2\epsilon[/tex]
Whereas [tex]\epsilon[/tex] accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:
[tex]0=f_{CH_4}^{in}-\epsilon \\\\\epsilon=30gmol/s[/tex]
Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:
[tex]f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_{CO_2}^{out}=15gmol/s\\\\f_{H_2O}^{out}=2*15gmol/s=30gmol/s[/tex]
It means that the mole fraction of carbon dioxide in that output is:
[tex]x_{CO_2}^{out}=\frac{15}{15+15+30} =0.25[/tex]
Best regards.
What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.
Answer:
i
[tex]J_{m} = 20 [/tex]
ii
[tex]J_{m} = 22.5 [/tex]
Explanation:
From the question we are told that
The first temperatures is [tex]T_1 = 25^oC = 25 +273 =298 \ K[/tex]
The second temperature is [tex]T_2 = 100^oC = 100 +273 = 373 \ K[/tex]
Generally the equation for the most highly populated rotational energy level is mathematically represented as
[tex]J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}[/tex]
Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]
Also
B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]
Generally the energy require per mole to move 1 cm is 12 J /mole
So [tex]0.244 \ cm^{-1}[/tex] will require x J/mole
[tex]x = 0.244 * 12[/tex]
=> [tex]x = 2.928 \ J/mol [/tex]
So at the first temperature
[tex]J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 20 [/tex]
So at the second temperature
[tex]J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 22.5 [/tex]
A bicycle rider is applying a force of 20 N while heading south against a wind blowing from the south with a force of 5 N
Answer:
the bike will go down 5 N
On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential energy of two O atoms versus the distance between their nuclei.
Answer:
Explanation:
We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.
From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.
Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.
which angles are right
Answer:
a right is 90 degrees
HELP FAST PLZ!!!!! Which phase change allows a substance to transform from a liquid to a
gas?
melting
Ofreezing
O ionization
condensation
deionization
O
evaporation
sublimation
Answer:evaporation
Explanation:
Answer:
Pretty sure its evaporation, if its not I'm very sorry.
The ion with the smallest diameter is ________. The ion with the smallest diameter is ________. Be2 Sr2 Ca2 Ba2 Mg2
Answer:
Be2^+
Explanation:
Ionic diameter increases down the group. This implies that Be2^+ will have the smallest diameter.
This extremely small diameter makes Be2^+ to differ considerably from other ions of group 2 elements.
For instance, the compounds of beryllium are mostly covalent in nature.
9th grade science need help ASAP
Answer:
D
Explanation:
It is a solid everything that is a solid is heavier
An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant. B) an increase in reaction rate due to a decrease in the kinetic energy of the reactants. C) an increase in the rate of reaction because reactant molecules collide with greater energy. D) an increase in both the reaction rate and activation energy due to increased kinetic energy.
Answer:
C) an increase in rate of reaction because reactant molecules collide with greater energy
Explanation:
Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.
The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.
When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq) Ag+ (aq) + I- (aq) → AgI (s) AgNO3 (aq) + KI (aq) → AgI (aq) + KNO3 (s) Ag+ (aq) + NO3 - (aq) → AgNO3 (aq) Ag+ (aq) + NO3 - (aq) → AgNO3 (s)
Answer:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Explanation:
We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.
Aqueous AgNO₃ and KI will dissociate in solution as follow:
AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)
KI (aq) —> K⁺(aq) + I¯(aq)
Aqueous AgNO₃ and KI will react as follow:
AgNO₃ (aq) + KI (aq) —>
Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)
Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.
Thus, the net ionic equation is:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
The net ionic equation of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex] to form [tex]AgI[/tex] precipitates is: B. [tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]
A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.
This ultimately implies that, both the charge on each atom and sum of the masses of the chemical compounds or elements in a chemical equation are properly balanced.
An ion can be defined as an atom or molecules (group of atoms) that has lost or gained one or more of its valence electrons, thereby, making it have a net positive or negative electrical charge.
First of all, we would write the dissociation equation for aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:
For [tex]AgNO_3[/tex]:
[tex]AgNO_3_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]
For [tex]KI[/tex]:
[tex]KI_{(aq)}[/tex] -----> [tex]K^{+}_{(aq)} + I^{-}_{(aq)}[/tex]
Next, we would write a chemical equation for the reaction of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:
[tex]AgNO_3_{(aq)} + KI_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex] [tex]+ \;K^{+}_{(aq)} + I^{-}_{(aq)}[/tex] ----->[tex]AgI_{(s)} + K^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]
Note: Spectator ions refers to the ions that exist as a reactant and a product in a chemical equation because they are unchanged by the chemical reaction.
In this chemical reaction, the spectator ions are:
[tex]K^+[/tex][tex]NO_{3}^{-}[/tex]Finally, in order to obtain the net ionic equation, we would cancel out the two (2) spectator ions:
[tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]
Read more: https://brainly.com/question/13750908
4. What reagent would you predict to be in excess for reacting 7.50 mL of a 0.10M BaCl2 solution with 7.50 mL of 0.10M KIO3 solution
Answer : [tex]BaCl_2[/tex] reagent predict to be in excess.
Explanation : Given,
Concentration of [tex]BaCl_2[/tex] = 0.10 M
Volume of [tex]BaCl_2[/tex] = 7.50 mL = 0.0075 L (1 L = 1000 mL)
Concentration of [tex]KIO_3[/tex] = 0.10 M
Volume of [tex]KIO_3[/tex] = 7.50 mL = 0.0075 L
First we have to calculate the moles of [tex]BaCl_2[/tex] and [tex]KIO_3[/tex].
[tex]\text{Moles of }BaCl_2=\text{Concentration of }BaCl_2\times \text{Volume of }BaCl_2[/tex]
[tex]\text{Moles of }BaCl_2=0.10M\times 0.0075L=0.00075mol[/tex]
and,
[tex]\text{Moles of }KIO_3=\text{Concentration of }KIO_3\times \text{Volume of }KIO_3[/tex]
[tex]\text{Moles of }KIO_3=0.10M\times 0.0075L=0.00075mol[/tex]
Now we have to calculate the excess and limiting reagent.
The balanced equilibrium reaction will be:
[tex]BaCl_2+2KIO_3\rightleftharpoons Ba(IO_3)_2+2KCl [/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]KIO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]
So, 0.00075 moles of [tex]KIO_3[/tex] react with [tex]\frac{0.00075}{2}=0.000375[/tex] moles of [tex]BaCl_2[/tex]
From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]KIO_3[/tex] is a limiting reagent and it limits the formation of product.
Hence, [tex]BaCl_2[/tex] reagent predict to be in excess.
What is the mass in grams of 2.5 moles of Al?
Answer:
One mole of Al weighs 27g.
2.5 moles of Al weigh 67.5g.
How many carbon atoms are in vitamin c?
Answer:
molecules can be much bigger. one molecule of vitamin c is made up of 20 atoms (6 carbons, 8 hydrogens, and 6 oxygens