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Mashcka [7]
3 years ago
5

Can anyone help ASAP!!

Chemistry
1 answer:
Sliva [168]3 years ago
4 0

The formula to determine the average atomic mass is:

Average atomic mass = mass of the isotope 1 \times natural abundance of isotope 1 + mass of the isotope 2 \times natural abundance of isotope 2 + ........ + mass of the isotope n \times natural abundance of isotope n -(1)

The atomic mass of first isotope of magnesium = 24 u  (given)

The natural abundance of first isotope = 78.70 %        (given)

The atomic mass of second isotope of magnesium = 25 u  (given)

The natural abundance of first isotope = 10.13 %        (given)

The atomic mass of third isotope of magnesium = 26 u  (given)

The natural abundance of first isotope = 11.17 %        (given)

Substituting the values in formula (1):

atomic weight = (24 \times 0.7870) + (25 \times 0.1013) + (26 \times 0.1117)

atomic weight = 24.4625 u

Hence, the atomic weight of naturally occurring isotopic mixture is 24.4625 u.


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In a calorimetry experiment 2.50 g of methane is burnt in excess oxygen. 30% of the energy released during the combustion is abs
Vikentia [17]

The total energy released per gram of methane burnt is  119,941.3 J/g.

<h3>Energy absorbed by water</h3>

Q = mcΔθ

where;

  • m is mass of water
  • c is specific heat of water
  • Δθ is change in temperature

Q = (500)(4.184)(68 - 25)

Q = 89,956 J

<h3>Total energy released per gram of methane burnt</h3>

0.3T =  89,956 J

T =  89,956 J/0.3

T = 299,853.3 J

Total energy per gram of methane, E = T.E/m

E = (299,853.3 J) / (2.5 g)

E = 119,941.3 J/g

Thus, the total energy released per gram of methane burnt is  119,941.3 J/g.

Learn more about energy here: brainly.com/question/25959744

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8 0
2 years ago
Why do carbs and fats have the same atoms, but store different amount of energy?
cestrela7 [59]
Carbs have larger molecules
7 0
3 years ago
Which of the following places the elements in the correct order of increasing first ionization energy?
Papessa [141]
A) Ca, Mg, Be
The first ionization energy increases as we move up a group because of the smaller atomic size and less shielding of electrons.
4 0
3 years ago
Read 2 more answers
Can anyone help me with this asap?
tresset_1 [31]

Answer:

Explanation:

1) During the diagnosis of thyroid disease a 10 g sample of I-131 is used. After a period of 32 days, how much sample is still radio active.?

Answer:

0.625 g

Explanation:

HL = Elapsed time/half life

32 days/8 days = 4

At time zero = 10 g

At 1st half life = 10/2 = 5 g

At 2nd half life = 5/2 = 2.5 g

At 3rd half life = 2.5 /2 = 1.25 g

At 4th half life = 1.25 / 2 = 0.625 g

After 32 days still 0.625 g of I-131 remain radioactive.

2) what was the original mass of sample Tc-99 that was used to locate the brain tumor If 0.10 g of a sample remains after 30 days? (half life 6 days)

Answer:

0.32 g.

Explanation:

Half life = time elapsed / HL

Half life = 30 days / 6 days = 5

At 5th half life = 0.10 g

At 4th half life = 0.2 g

At 3rd half life =  0.4 g

At 2nd  half life = 0.8 g

At 1st half life = 0.16 g

At time zero = 0.32 g

The original amount was 0.32 g.

3) write the beta decay equation of I-131?

Equation:

¹³¹I₅₃  →  ¹³¹Xe₅₄ + ⁰₋₁e

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiation

¹³¹I₅₃  →  ¹³¹Xe₅₄ + ⁰₋₁e

The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.

3 0
3 years ago
Caluclate the volume of0,25 mol dm -^3 hydrochloric acid required to neutralize 2.0 dm^3 of 0,15 mol dm-^3 barium hydroxide
inessss [21]

Answer:

2.4 dm-3

Explanation:

Equation of the reaction;

2HCl(aq) + Ba(OH)2(aq) -------> BaCl2(aq) + 2 H2O(l)

Volume of acid VA = ??

Concentration of acid CA = 0.25 moldm-3

Volume of base VB = 2.0 dm^3

Concentration of base CB = 0.15 moldm-3

Number of moles of acid NA = 2

Number of moles of base NB = 1

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VA = CBVBNA/CANB

VA = 0.15 * 2 * 2/0.25 *1

VA = 2.4 dm-3

4 0
3 years ago
Read 2 more answers
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