How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g.

Question: How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g. Use correct number of significant digits;

Solution: Mol weight of Ca(NO3)2 = 90 + 28 +96
= 164 g/mol
100 g of Ca(NO3)2 = 0.6097 mol of Ca(NO3)2
1 mol of Ca(NO3)2 contains 6 number of oxygen atoms.
so, 0.6097 moles of Ca(NO3)2 contains = 0.6097 * 6 = 3.6582 Now, mole of each ‘O’ atom is 16 g/mol So, 100 g of Ca(NO3)2 contains = 3.6582*16 = 58.5312 g of oxygen.

A superconductor is a substance that is able to conduct electricity without resistance

Question: A superconductor is a substance that is able to conduct electricity without resistance, a property that is very desirable in the construction of large electromagnets. Metals have this property if cooled to temperatures a few degrees above absolute zero, but this requires the use of expensive liquid helium (boiling point 4 K). Scientists have discovered materials that become superconductors at higher temperatures, but they are ceramics. Their brittle nature has so far prevented them from being made into long wires. A recently discovered compound of magnesium and boron, which consists of 52.9% Mg and 47.1% B, shows special promise as a high-temperature superconductor because it is inexpensive to make and can be fabricated into wire relatively easily. What is the formula of this compound?

Solution:

Let us take 100 g of compound
Then, mass of Mg = 52.9g
and mass of boron = 47.1g
Moles of Mg = Mass/Molar Mass = 52.9g/24.31g/mol
= 2.18 mol
and moles of B = 47.1g/10.81g/mol
= 4.36 mol
So, the molar ratio of Mg and B:
nMg/nB = 2.18 mol/4.36 mol = 1/2
Formula of the compound is MgB2

Assuming that all the blackbody radiation is emitted at this wavelength

Question: Consider a metal that is being welded

Assuming that all the blackbody radiation is emitted at this wavelength (a crude approximation), how many photons does a square centimeter of the metal emit per second?

Solution: Using stefan-boltzmann law, energy emitted by blackbody
E = σAtT^4
A = 1 cm^2 = 10^-4 m^2
t = 1s
but E =hcn/λ
σ = 5.67 * 10^-8
T = b/λ = 2.898 * 10^-3/λ
n = λσAtT^4/hc
n = 5.6710^-810^-41(2.89810^-3)^4/(6.6310^-34310^8(49010^-9)^3)
n= 1.71*10^22 photons

Drag the “orbitals with electrons” box to the energy scale shown.

Question: Drag the “orbitals with electrons” box to the energy scale shown. Your goal in this exercise is to show nitrogen in an sp3-hybridized state which is capable of forming 3 bonds with three other atoms. Fill answer fields from left to right. Place orbitals with two electrons on the left (if any).

Solution: Concepts and reason

An orbital is the distribution of the probability density that is associated with the given electron centered at a given point. If the point is the nucleus of an atom, an atomic orbital is the resultant. If the point is the center of mass of multiple nuclei within a molecule, a molecular orbital is the resultant.

When an atom is required to bond with the other atoms to form a molecule, the valence atomic orbitals with almost similar energies ‘hybridize’ to form hybridized orbitals. These orbitals of identical shapes have identical associated energies.

The number of hybrid orbitals formed is equal to the number of atomic orbital hybridizing. Fundamentals

The hybridization of an atom in a molecule could be correlated with the number of bonds formed by it. This is tabulated as follows:

The sum of the principal quantum number and the azimuthal quantum number informs about the energy associated with a given orbital.

(n+l)\left( {{\rm{n + l}}} \right)(n+l)

Here

n−{\rm{n}}\,{\rm{ - }}\,n− Principal quantum number

l−{\rm{l}}\,{\rm{ - }}l− Azimuthal quantum number

The electronic configuration of nitrogen is 1s22s22p3{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{3}}}1s22s22p3 .

The (n+l)\left( {{\rm{n + l}}} \right)(n+l) values associated with the atomic orbitals involved are as follows:

1s−12s−22p−3\begin{array}{l}\\{\rm{1s}}\,{\rm{ - }}\,{\rm{1}}\\\\{\rm{2s}}\,{\rm{ - }}\,2\\\\{\rm{2p}}\,{\rm{ - }}\,{\rm{3}}\\\end{array}1s−12s−22p−3​

The atomic orbitals with lower (n+l)\left( {{\rm{n + l}}} \right)(n+l)​ values have lower associated energies. Therefore, they are filled first. In nitrogen, the orbital with the lowest associated energy, 1s{\rm{1s}}1s​ , is filled foremost, followed by 2s{\rm{2s}}2s​ , and finally ​2p{\rm{2p}}2p ​.

The hybridized orbitals possess a blend of characteristics associated with their constituent atomic orbitals. Therefore, they could be placed at an intermediate position between the respective atomic orbitals (here, 2sand2p{\rm{2s}}\,{\rm{and}}\,{\rm{2p}}2sand2p ).

Hence, the arrangement is given below:

The arrangement is given below:

The sp3{\rm{s}}{{\rm{p}}^{\rm{3}}}sp3 orbitals possess 25%{\rm{25\% }}25% ‘s’ character and 75%{\rm{75\% }}75% ‘p’ character. Correspondingly, the stabilization or the effective energy of the hybridized orbitals is a little lesser than the un-hybridized ′2p′{\rm{'2p'}}′2p′ orbitals, but quite greater than the un-hybridized ′2s′{\rm{'2s'}}′2s′ orbitals. The electronic population of the un-hybridized, valence orbitals is 666 . Correspondingly, the hybridized orbitals also have the same population.

Create a class called Invoice that a hardware store might use to represent an invoice for an item sold at the store.

Question: Create a class called Invoice that a hardware store might use to represent an invoice for an item sold at the store. An Invoice should include four pieces of information as instance variables — a part number(type String), a part description(type String), a quantity of the item being purchased (type Integer) and a price per item (type Integer). Your class should have a constructor that initializes the four instance variables. Provide a property for each instance variable. If the quantity is not positive, it should be set to 0. If the price per item is not positive, it should be set to 0; Use validation in the properties for these instance variables to ensure that they remain positive. In addition, provide a method named DisplayInvoiceAmount that calculates and displays the invoice amount (that is, multiplies the quantity by the price per item). How do you write an application that demonstrates the class Invoice’s capabilities?

Solution:

import java.util.Scanner;

class Invoice{

String partNumber;

String partDescription;

int itemPurchased;

double pricePerItem;

Invoice(){

partNumber = “”;

partDescription = “”;

itemPurchased = 0;

pricePerItem = 0.0;

}

String getPartNumber(){

return partNumber;

}

String getPartDescription(){

return partDescription;

}

int getItemPurchased(){

return itemPurchased;

}

double getPricePerItem(){

return pricePerItem;

}

double getInvoiceAmount(){

return (itemPurchased * pricePerItem);

}

void setPartNumber(String pn){

partNumber = pn;

}

void setPartDescription(String pd){

partDescription = pd;

}

void setItemPurchased(int ip){

itemPurchased = ip;

}

void setPricePerItem(double ppi){

pricePerItem = ppi;

}

}

class InvoiceDemo {

public static void main(String args[]) {

Scanner sc = new Scanner(System.in);

Invoice invoice = new Invoice();

System.out.print(“\nEnter part number :”);

invoice.setPartNumber(sc.nextLine());

System.out.print(“Enter part description :”);

invoice.setPartDescription(sc.nextLine());

System.out.print(“Enter item purchased :”);

invoice.setItemPurchased(sc.nextInt());

System.out.print(“Enter price per item :”);

invoice.setPricePerItem(sc.nextDouble());

System.out.print(“\n\nDetail of items purchasing–>”);

System.out.print(“\nPart number :” + invoice.getPartNumber());

System.out.print(“\nPart description :” + invoice.getPartDescription());

System.out.print(“\nTotal Billing Amount :” + invoice.getInvoiceAmount());

}

Use input() function to ask the users to input the courses and professors in this semester

Question: Use input() function to ask the users to input the courses and professors in this semester, until the users input “done” as an end. Put the courses and the professor names into two separate lists and print out these lists as follows (don’t print lists literally). Then print out the following statements using for loop in each list. Please note there are no newlines in the first statement, while there are ‘->‘ symbols in the second statement. You cannot print the statements literally.

Solution:


course_list = []
professor_list = []
while(True):
    course = input("Please enter the courses you are taking in this semester :")
    if(course=="done"):
        break;
    professor = input("Please enter the professor's name for the course:")
    course_list.append(course)
    professor_list.append(professor)
   
print("The list of courses is : ", course_list)
print("The list of Professors is: ", professor_list)
print("The course you are taking int this semester are:",end=" ")
for i in course_list:
    print(i,end=",")
print()
print("The profesor of these courses are ")
for i in professor_list:
    print(str(i)+"->")

Output:

Discrete Example Suppose we have a 4-sided die and let X denote the random face that comes up on a throw. Its pmf is given by Table 1, where θ,p ∈ [0,1]. Suppose we throw

Question: Discrete Example Suppose we have a 4-sided die and let X denote the random face that comes up on a throw. Its pmf is given by Table 1, where θ,p ∈ [0,1]. Suppose we throw

X1234
Px(x)θp(1-θ)pθ(1-p)(1-θ)(1-p)
Table 1: Pmf of X

the die a certain number of times and observe xi i’s, for i = 1,…,4 (i.e., face i comes up xi times).
(a) What is the likelihood of this experiment given θ? (You should treat p as a constant) (b) What is the maximum likelihood estimate of θ?

 3.2 Normal Distribution

 Suppose we sample n i.i.d. points from a Gaussian distribution with mean μ and variance σ?. Recall that the Gaussian pdf is given by

Compute the maximum likelihood estimate of parameters μ and σ?.

Solution:

Suppose that a van with capacity for 7 passengers departs from a commuter station

Question: Suppose that a van with capacity for 7 passengers departs from a commuter station. Observation has shown that the bus never departs empty. Let X denote the number of passengers that are female and Y the number of passengers that are male. Assume that all possible (x, y) pairs are equally likely. (a) Find the probability that the number of female passengers is more than 5. (a) Find the probability that the total number of passengers is less than 5.

Solution:

pairs of passengers:-
{(0,7),(0,6),(0,5),(0,4),(0,3),(0,2),(0,1)
(1,6),(1,5),(1,4),(1,3),(1,2),(1,1),(1,0)
(2,5),(2,4),(2,3),(2,2),(2,1),(2,0)
(3,4),(3,3),(3,2),(3,1),(3,0)
(4,3),(4,2),(4,1),(4,0),
(5,2),(5,1),(5,0)
(6,1),(6,0)
(7,0)}
p(x,y) = 1/35 where x = no of female and y = no of males

a. Probability of female passengers is more than 5
possible outcomes = (6,0),(6,1),(7,0)
P(A) = 3/35 = 0.0857

b. Passenger is less than 5
possible outcomes = {(4,0),(3,0),(3,1),(2,2),(2,1),(2,0),
(1,3),(1,2),(1,1),(1,0),(0,3),(0,2),(0,1),(0,4)}
= 14
P(B) = 14/35 = 0.4

Which statement below about asexual reproduction is FALSE?

Question: Which statement below about asexual reproduction is FALSE?
A. With asexual reproduction, offspring are genetically equivalent to the parent.
B. Asexual reproduction requires no partner.
C. Asexual reproduction requires meiosis.
D. Some organisms can reproduce both sexually and asexually.

Solution: Asexual reproduction requires meiosis.
only sexual reproduction required meiosis, as gamete fusion occurs, to maintain the stable chromosome number in the offspring due to this phenomenon, meiosis is required in sexual reproduction.

Question: If a diploid sperm fertilized a diploid egg, what would the result be?
A. Diploid (2n)
B. Quadruploid (4n)
C. A mixture of diploid (2n) and quadruploid (4n)
D. Triploid (3n)

Solution: The answer is Quaadruploid (4n)
2n+2n = 4n.

Question: Which of the following represents things that are equivalent?
A. Two alleles for the same gene in a homologous chromosome pair
B. The sequences of DNA in the two sister chromatids of a chromosome before meiosis
C. The sequences of DNA in the two sister chromatids of a chromosome after meiosis
D. The number of homologous chromosomes per cell before and after meiosis

Solution: Two alleles for the same gene in a homologous chromosome pair.

Question: What problem would most likely occur if a haploid cell attempted to perform meiosis?
A. The cell could not replicate its DNA prior to meiosis.
B. The cell could not pair homologous chromosomes during meiosis I.
C. The cell would produce diploid daughter cells.

Solution: The cell could not pair homologous chromosomes during meiosis I.

Question: Which statement below is TRUE?
A. Crossing over shuffles genes between non-homologous chromosomes.
B. During melosis, cells arrange their alleles so that beneficial alleles are passed on together.
C. Without crossing over, offspring would be genetically identical to parents.
D. New combinations of alleles arise from the random events of crossing over, independent assortment, and mutation.

Solution: New combinations of alleles arise form the random events of crossing over, independent assortment, and mutation.

Question: What statement below is TRUE of the spindle during mitosis and meiosis?
A. The spindle always separates sister chromatids during anaphase.
B. The spindle always separates homologous chromosomes during anaphase.
C. Chromosomes are always attached to both spindle poles during metaphase.
D. The spindle always attaches to chromosomes at the kinetochore.

Solution: The spindle always attached to chromosomes at the kinetochore.

Question: If a person is heterozygous for the △32 allele of the CCR5 gene, how many of the four daughter cells produced by meiosis will have the △32 allele?
A. 1
B. 2
C. 4
D. Varies

Solution: 2.

Question: Which of the following processes has the effect of changing ploidy?
A. DNA replication
B. Meiosis I
C. Meiosis II
D. Mitosis

Solution: Meiosis I
During Anpahse I of meiosis I, as the chromosome segregation occurs, the ploidy will be reduced to half.

Question: The image below shows a homologous chromosome pair with two genes, A and B. Each of the two genes has two alleles (A1 and A2, B1 and B2). For this homologous pair of chromosomes, after completion of meiosis with NO CROSSING OVER, which of the following combinations of alleles might end up in one of the gametes?


A. A1 and B1
B. A2 and B1
C. A2 and B2
D. All of the above are possible.

Solution: A2 and B1

Question: For the same homologous pair of chromosomes shown in the previous question, which of the following combinations of alleles might end up in one of the gametes after completion of meiosis WITH CROSSING OVER?
A. A1 and B2
B. A2 and B1
C. A2 and B2
D. All of the above are possible

Solution: All of the above are possible.