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pantera1 [17]
3 years ago
8

an isotope or barium (atomic number 56)has an mass of 138.how many neutrons are in the nucleus of this isotope

Chemistry
1 answer:
tia_tia [17]3 years ago
8 0

Answer:

Diagram of the nuclear composition and electron configuration of an atom of barium-138 (atomic number: 56), the most common isotope of this element. The nucleus consists of 56 protons (red) and 82 neutrons (blue).

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An artifact contains one-fourth as much carbon-14 as the atmosphere. how old is the artifact
Nikolay [14]
<h3><u>Answer;</u></h3>

= 11,460 years

<h3><u>Explanation;</u></h3>
  • <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass.  Therefore, it  would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>

<em>The initial amount of carbon-14 in this case was 1 whole; thus; </em>

<em>1 → 1/2 →1/4</em>

<em>To contain 1/4 of the value, 2 half-lives have passed. </em>

<em>But, 1 half life = 5,730 years</em>

<em>Therefore; The artifact is is therefore: 2 x 5,730 </em>

<em>          = 11,460 years </em>

4 0
3 years ago
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The table below shows the electronegativity values of various elements on the periodic table. Electronegativities A partial peri
MatroZZZ [7]

Answer:

Oxygen and Chlorine

Explanation:

Covalent bonds involve the sharing of electrons between nonmetals.

4 0
4 years ago
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The actions described in the passage are best understood in the context of which of the following?
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6 0
3 years ago
if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?
ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

<em>(XMV) acid = (XMV) base.</em>

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HI = (XMV) Ca(OH)₂.</em>

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>

6 0
4 years ago
7.296 x 10^2 / 9.6 x 10^-9
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