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icang [17]
2 years ago
6

Many hospitals, and some doctors\' offices, use radioisotopes for diagnosis and treatment, or in palliative care (relief of symp

toms such as pain). Some radioisotopes used in medicine are listed below. Write the isotope symbol for each radioisotope. Replace the question marks with the proper integers. Replace the letter X with the proper element symbol.
a) Iodine-131:

b) Iridium-192:

c) Samarium-153:
Chemistry
1 answer:
Phantasy [73]2 years ago
8 0

Answer:

a) ^{131}_{53} I

b) ^{192}_{77} Ir

c) ^{153}_{62} Sm

Explanation:

The symbols of the isotopes are written like

^{A}_{Z} X

where,

X is the element

A is the mass number (protons + neutrons)

Z is the atomic number (protons)

<em>a) Iodine-131</em>

The atomic number of iodine is 53. The mass number of this isotope is 131. The symbol is ^{131}_{53} I.

<em>b) Iridium-192</em>

The atomic number of iridium is 77. The mass number of this isotope is 192. The symbol is ^{192}_{77} Ir.

<em>c) Samarium-153</em>

The atomic number of samarium is 62. The mass number of this isotope is 153. The symbol is ^{153}_{62} Sm.

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Answer:

No, gravity isn't matter

Explanation:

Gravity is a <u>force</u> that attracts matter towards the center of a physical body with mass.

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GIVING BRAINLIEST!!! A student observes a beaker of room temperature water resting on a table. She states that the beaker of wat
defon

Answer:

a

Explanation:

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Read 2 more answers
How many molecules of propane were in the erlenmeyer flask? Avogadro's number is 6. 022 × 10^23 molecules/mol
Ede4ka [16]

3.74×10^{21}

3.74 ×10^{21} molecules of propane were in the erlenmeyer flask.

number of moles of propane can be calculated as moles of propane.

mass of propane =  0.274 g

molar mass of propane = 44.1

So this gives us the value of 6.21×10^{-3} moles of propane

No one mole of propane As a 6.0-2 × 10^{23}

so, 6.21 ×10^{-3} × 6. 022 × 10^23

= 3.74 ×10^{21}

Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 ×10^{21}

<h3>What is erlenmeyer flask?</h3>
  • A laboratory flask with a flat bottom, a conical body, and a cylindrical neck is known as an Erlenmeyer flask, sometimes known as a conical flask or a titration flask.
  • It bears the name Emil Erlenmeyer after the German chemist.

<h3>What purpose does an Erlenmeyer flask serve?</h3>
  • Liquids are contained in Erlenmeyer flasks, which are also used for mixing, heating, chilling, incubating, filtering, storing, and other liquid-handling procedures.
  • For titrations and boiling liquids, their sloped sides and small necks make it possible to whirl the contents without worrying about spills.

To learn more about calculating total molecules visit:

brainly.com/question/8933381

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4 0
1 year ago
Why does beryllium lose electrons when forming ionic bonds,while sulfur gains electrons?
Nadya [2.5K]
Beryllium is a metal so it needs to loose electrons to have a full outer shell while sulfur is a non-metal so it needs to gain electrons to fill its outer shell.
8 0
3 years ago
Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with
VladimirAG [237]

Answer:

7.3 g (NH₄)₃PO₄

Explanation:

The balanced equation for the reaction is:

H₃PO₄ +  3 NH₃ ----> (NH₄)₃PO₄

To find the mass of ammonium phosphate ((NH₄)₃PO₄) produced, you need to (1) convert grams NH₃ to moles NH₃ (via the molar mass from the periodic table), then (2) convert moles NH₃ to moles (NH₄)₃PO₄ (via mole-to-mole ratio from balanced equation), and then (3) convert moles (NH₄)₃PO₄ to grams (NH₄)₃PO₄ (via molar mass from periodic table). Make sure to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs because the given value (2.5 grams) has 2 sig figs.

Molar Mass (NH₃): 14.01 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.034 g/mol

Molar Mass ((NH₄)₃PO₄):

3(14.01 g/mol) + 12(1.008 g/mol) + 30.97 g/mol + 4(16.00 g/mol)

Molar Mass ((NH₄)₃PO₄): 149.096 g/mol

2.5 g NH₃          1 mole NH₃        1 mole (NH₄)₃PO₄             149.096 g
---------------  x  --------------------  x  ---------------------------  x  --------------------------
                            17.034 g             3 moles NH₃              1 mole (NH₄)₃PO₄

=  7.3 g (NH₄)₃PO₄

8 0
1 year ago
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