answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GalinKa [24]
2 years ago
8

Calculate the mass defect of the boron nucleus 11 5b. The mass of neutral 11 5b is equal to 11.009305 atomic mass units.

Chemistry
2 answers:
yan [13]2 years ago
8 0

The mass defect of the Boron : <u>Δm = 0.081815 amu</u>

<h3>Further explanation</h3>

Atoms are composed of 3 types of basic particles namely protons, electrons, and neutrons

The mass of this basic particle is expressed in atomic mass units (amu)

The mass of atoms is always slightly less than the sum of the masses of the individual atomic forming particles namely neutrons, protons, and electrons. The difference between the mass of the atom and the sum of the masses of atomic subparticles is called the mass defect (Δm).

Can be formulated:

Δm = [Z (mp + me) + (A-Z) mn] - m atom

where:

Δm = mass defect (amu)

mp = mass of a proton (1,007277 amu)

mn = mass of a neutron (1.008665 amu)

me = mass of an electron (0.000548597 amu)

matom = mass of nuclide (amu)

Z = atomic number

A = mass number

Atomic number (Z) = number of protons = number of electrons

Mass Number (A) is the sum of protons and neutrons

Atomic Number (Z) = Mass Number (A) - Number of Neutrons

For Boron elements (₁₁⁵B) which have atomic number 5 and mass number 11, then:

number of protons = 5

number of electrons = 5

number of neutrons = mass number - atomic number = 11 - 5 = 6

then:

Δm = [5 (1.007277 + 0.000548597) + (6) 1.008665] - 11.009305

Δm = [5 (1.007826) + (6) 1.008665] - 11.009305

Δm = (5.03913 + 6.05199) - 11.009305

Δm = 11.09112 - 11.009305

Δm = 0.081815 amu

<h3>Learn more </h3>

element symbol

brainly.com/question/2572495

the chemical symbols

brainly.com/question/8698247

The subatomic particle that has the least mass  

brainly.com/question/2224691  

Keywords: mass number, atomic mass, amu, mass defect, atomic number

Vlad1618 [11]2 years ago
7 0

Answer:

0.07906687 amu

Explanation:

For Boron ₅B¹¹, the number of protons is 5 and the mass is 11. The mass is the number of protons plus the number of neutrons, so:

neutrons = 11 - 5 = 6

The mass of an atom is concentrated in the nucleus, so it is the mass of the protons + the mass of the neutrons. The mass of 1 proton is 1.00727647 amu/proton, and the mass of 1 neutron: 1.00866492 amu/neutron, so for the element given the theoretical mass (mt) is:

mt = 5* 1.00727647 amu/proton + 6*1.00866492 amu/neutron

mt = 11.08837187 amu

The mass defect (md) is the theorical mass less the real mass:

md = 11.08837187 - 11.009305

md = 0.07906687 amu

You might be interested in
A sample of an unknown substance has a mass of .158 kg if 2,510 J of heat is required to heat the substance from 32°C to 61°C wh
ziro4ka [17]

Answer: 0.548J/g°C

Explanation:

Q = s × m × DeltaT

Q = Heat (J)

S = Specific Heat Capacity

M = mass (g)

DeltaT = Change in temperature (°C)

0.158Kg x 1000 = 158g

2.510J = s x 158g x (61°C-32°C)

2.510J/(158g x 29°C) = s

S = 0.54779.... J/g°C

S = 0.548 J/g°C

7 0
2 years ago
A 52.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 26.0 mL of KOH at 25 ∘C.
prohojiy [21]
The balanced equation for the above reaction is 
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL  = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs. 
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7
 
7 0
2 years ago
If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g)
Allisa [31]

Answer:

The correct answer would be : 33.8 g

Explanation:

Molar mass of ammonia,

Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)

= 1*14.01 + 3*1.008  = 17.034 g/mol

mass(NH3)= 25.0 g  (given)

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(25.0 g)/(17.034 g/mol)

= 1.468 mol

Now,

Molar mass of O2

= 32 g/mol

mass(O2)= 45.0 g

similar as ammonia

n (O2)=(45.0 g)/(32 g/mol)

= 1.406 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

1.83456 mol of O2 is required  for 1.46765 mol of NH3

by the calculation we have only 1.40625 mol of O2

Thus, the limiting agent will be - O2

now the Molar mass of NO,

= 1*14.01 + 1*16.0

= 30.01 g/mol  (similar formula used for NH3)

Balanced equation :

mol of NO formed = (4/5)* moles of O2

= (4/5)×1.40625  (from above calculation)

= 1.125 mol

mass of NO = number of moles × molar mass

= 1.125*30.01

= 33.8 g

Thus, the correct answer would be : 33.8 g

5 0
2 years ago
A gas has a pressure of 3.16 atm at STP. I have decided to transfer it to a container that is 3 times larger than the original v
Maru [420]

Answer: The new pressure will be 1.1atm

Explanation:

V1 = V

P1 = 3.16atm

V2 = 3V

P2=?

P1V1 =P2V2

3.16 x V = P2 x 3V

P2 = (3.16 x V) /3V

P2 = 1.1atm

7 0
2 years ago
A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m
ioda
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = \frac{PV}{nT} = constant
\frac{P1 V1}{n T} = \frac{P2 V2}{n T}
==> P1 * V1 = P2 * V2
V2 = \frac{P1 V1}{P2} = \frac{445 0.016}{1225}
V2 = 0.00581 m3 = 5.81 L


7 0
2 years ago
Other questions:
  • The average kinetic energy of water molecules is greatest in which of these samples?(1) 10 g of water at 35°C(2) 10 g of water a
    14·2 answers
  • Lewis structure XeH4? The hydrogen atom is not actually electronegative enough to form bonds to xenon. Were the xenon-hydrogen b
    6·2 answers
  • The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) calculate kp for the reaction below.
    15·2 answers
  • A radioisotope is placed near a radiation detector, which registers 64 counts per second. Eight hours later, the detector regist
    11·2 answers
  • Write a balanced chemical equation for the reaction of copper(II) sulfate and concentrated ammonia to produce teramine copper(II
    5·1 answer
  • The reaction A  B + C is second order in A. When [A]0 = 0.100 M, the reaction is 35.0% complete in 32.3 hours. Calculate the va
    15·1 answer
  • Question 13 A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The con
    14·1 answer
  • A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc
    10·1 answer
  • A student in a chemistry laboratory has access to two acid solutions. The first one is 20% acid and the second solution is 45% a
    12·1 answer
  • 9 The Haber process is a reversible reaction. N2(g) + 3H2(g) 2NH3(g) The reaction has a 30% yield of ammonia. Which volume of am
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!