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barxatty [35]
3 years ago
9

How is radioactive elements used for medical diagnosis

Chemistry
1 answer:
Lynna [10]3 years ago
5 0
It is all of the above.
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Sooeme and I just wanted you guys know how much I appreciate you guys wanna offer me and I have your
Marat540 [252]

Answer:

Your answers are solids, liquids, and gases.

Explanation:

These are the three states of matter.

<em>Please</em><em> like</em><em> and</em><em> mark</em><em> brainliest</em><em>!</em>

<em>Hope</em><em> it</em><em> helps</em><em>!</em>

4 0
3 years ago
Read 2 more answers
3. If you start with 8x1025 molecules of Cl, and 25 grams of KI, how many grams of KCl would
miskamm [114]

Answer:

Percent yield = 89.1%

Explanation:

Based on the equation:

Cl₂ + 2KI → 2KCl + I₂

<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>

<em />

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

<em>Moles Cl₂:</em>

8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles

<em>Moles KI -Molar mass: 166.0028g/mol-</em>

25g * (1mol / 166.0028g) = 0.15 moles

Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

11.2g KCl

Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

<h3>Percent yield = 89.1%</h3>
3 0
3 years ago
The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?
vazorg [7]
C + O2= CO2
n  =  \frac{m}{mw}
n =  \frac{5.4}{12}  \\ n  = 0.45 \: mol \: of \: carbon
n =  \frac{13.6}{12 + 16 \times 2} \\ n =  \frac{13.6}{44}  \\ n = 0.31 \: mol \: of \: carbon \: dioxide
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
4 0
3 years ago
Calculate the mole fraction of cai2 in an aqueous solution prepared by dissolving 0.400 moles of cai2 in 850.0 g of water.
alukav5142 [94]
1) Formulas:

a) mole fraction of component 1, X1

X1 =  number of moles of compoent 1 / total number of moles

b) Molar mass = number grams / number of moles => number of moles =  number of grams / molar mass


2) Application

Number of moles of CaI2 = 0.400

Molar mass of water = 18.0 g/mol

Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol

Total number of moles = 0.400 + 47.22 =47.62

Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840
7 0
3 years ago
How many moles of oxygen would be needed to produce 84 moles of sulfur trioxide according to the following balanced chemical equ
olganol [36]

Answer:

126 moles

Explanation:

2S +3 o2=2so3

So if 2 moles of so3 required 3 moles of oxygen

. So 84 moles of so3 will require 84*3/2=126 moles of oxygen

4 0
3 years ago
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