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boyakko [2]
1 year ago
14

calculate the average atomic mass of lead based on the given data and the atomic masses. isotope atomic mass (amu) percent abund

ance pb-206 205.97 14.5 pb-207 206.98 17.0 pb-208 207.98 68.5
Chemistry
1 answer:
BaLLatris [955]1 year ago
8 0

The average atomic mass of lead given its three isotopes is 207.52.

<h3>What is average atomic mass?</h3>

Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element.

According to this question, the following isotopes with their percent abundance and atomic masses of lead are given:

  • Pb-206 - 205.97 14.5%
  • Pb-207 - 206.98 17.0%
  • Pb-208 - 207.98 68.5%

  • Pb-206 = 14.5/100 × 205.97
  • Pb-207 = 17/100 × 206.98
  • Pb-208 = 68.5/100 × 207.98

  • Pb-206 = 29.865
  • Pb-207 = 35.186
  • Pb-208 = 142.466

The average atomic mass is as follows: 29.865 + 35.186 + 142.466 = 207.52

Therefore, the average atomic mass of lead given its three isotopes is 207.52.

Learn more about average atomic mass at: brainly.com/question/13753702

#SPJ1

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A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s.
Semmy [17]

Answer

2.7956 * 10^19 photons

Givens

  • Wavelength = λ = 525 * 10^-9 meters  [1 nmeter = 1*10^-9 meters]
  • c = 3 * 10^8 meters
  • E = ???
  • W = 100  watts
  • t = 1 second
  • h= plank's Constant = 6.26 * 10^-34 J*s

Formula

E = h * c / λ

W = E / t

Solution

E = 6.26 * 10^-34 j*s * 3 * 10^8 m/s /525 * 10^-9 (m)

The meters cancel out. So do the seconds. You are left with Joules as you should be.

E =  3.577 * 10^-18 Joules

What you have found is the energy of 1 photon.

Now you have to find the Joules from the watts.

W = E/t

100 * 1 second = 100 joules

1 photon contains 3.577 * 10 ^ - 18 Joules

x photon = 100 joules                        

1/x = 3.577 * 10^-18 / 100                          Cross multiply

100 = 3.577 * 10 ^ - 18 * x                         Divide both sides by 3.577 * 10 ^ - 18

100/3.577 * 10 ^ - 18 = 3.577 * 10 ^ - 18x / 3.577 * 10 ^ - 18

2.7956 * 10^19 photons = x


7 0
3 years ago
Which of the following solutions is a good buffer system?A solution that is 0.10 M HCN and 0.10 M NaClA solution that is 0.10 M
patriot [66]

Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN

Explanation:

  • A good buffer system contains a weak acid and its salt or a weak base and its salt.
  • In this case; A solution that is 0.10 M HCN and 0.10 M LiCN, would make a good buffer system.
  • HCN is a weak acid, while LiCN is a salt of the weak acid, that is, CN- conjugate of the acid.
8 0
3 years ago
How many bromine atoms are present in 39.0 g of ch2br2?
Mandarinka [93]
Answer is: there is  2,69·10²³ atoms of bromine.
m(CH₂Br₂) = 39,0 g.
n(CH₂Br₂) = m(CH₂Br₂) ÷ M(CH₂Br₂).
n(CH₂Br₂) = 39 g ÷ 173,83 g/mol.
n(CH₂Br₂) = 0,224 mol.
In one molecule of CH₂Br₂, there is two bromine atoms, so:
n(CH₂Br₂) : n(Br) = 1 : 2.
n(Br) = 0,448 mol.
N(Br) = n(Br) · Na.
N(Br) = 0,448 mol · 6,022·10²³ 1/mol.
n(Br) = 2,69·10²³.
7 0
3 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
Give the orbital notation for a neutral atom of titanium
dedylja [7]

Answer:

Explanation:

A neutral titanium atom will have 22 electrons. Therefore, its electron configuration will be 1s2 2s2 2p6 3s2 3p6 4s2 3d2.

5 0
2 years ago
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