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irinina [24]
3 years ago
8

In a college of exactly 2760 students, exactly 65 % are male. What is the number of female students? Express your answer as an i

nteger.
Chemistry
1 answer:
lina2011 [118]3 years ago
5 0

Total number of students (male and female) in the college=2760. Total number of students including both male and female represents 100% (percentage) of students which is equal to 2760 number.

Number of male students is 65% (percentage) which is equal to \frac{2760X65}{100}= 1794. So, total number of female students are (100-65)%=35% (percentage) which is equal to \frac{2760X35}{100}=966 numbers.


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Why does a metal bond with a non-metal in ionic bondings?? ASAP please
crimeas [40]

An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between atoms.

4 0
3 years ago
What is the name of this alkane? two central carbons are bonded to c h 3 at each end, h below, and c h3 above the left carbon an
Kazeer [188]

The name of this alkane is with central carbons are bonded to c h 3 is 2-methylbutane.

<h3>What is alkane?</h3>

Alkanes belong to the family of  saturated hydrocarbons with carbon carbon single bond.

For the given alkane;

          CH₃    H

 CH₃ -  C   -  C - CH₃

            H      H

Thus, the name of this alkane is with central carbons are bonded to c h 3 is 2-methylbutane.

Learn more about alkane here: brainly.com/question/24270289

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4 0
2 years ago
The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,
Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

\frac {r^+}{r^-}=0.731 .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

r^++r^-=\frac {a}{2}  ...................2

Given that:

Cl^-\ (r^-) = 1.82\ \dot{A}

To find,

K^+\ (r^+) = ? \dot{A}

Using 1 and 2 , we get:

1.731\ r^+=0.731\times \frac {6.28}{2}

<u>Size of the potassium ion = 1.33 Å</u>

4 0
3 years ago
List the protons, neutrons, and electrons for Ge2+<br> Please hurry
Nady [450]

Answer:

32, 30 and 41

Explanation:

The problem here is to find the number of:

    Protons, neutrons and electrons in Ge²⁺

In this ion,

   We must understand that for a net positive charge to remain on an atom, the number of protons must be greater than the number of electrons.

Ge is Germanium with atomic number of 32;

So the number of protons is 32

Since the atom has lost two electrons;

  Number of electrons now is 32  - 2 = 30

Number of neutrons is 41 from the periodic table.

7 0
2 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
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