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sertanlavr [38]
2 years ago
7

Please help me with this three questions

Chemistry
1 answer:
zheka24 [161]2 years ago
7 0

Answer:

1. A. Thyroid Gland Function

B. detecting Blood Clots

C. Treating Cancer

2. An example would be how many are used as tracers for diagnostic purposes

3. Beneficial because it gives us ways to eleminate pests, by the sterile insect technique

Number three is kinda iffy a bit difficult to explain it may be explained better in your book.

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What is the indicator of the number of ions in solution?
Andru [333]

Answer:

The amount of solute added.

Explanation:

The amount of solute added is directly proportional to the number of ions.

The higher the amount added the higher the number of moles.

The number of moles is multiplied by the Avogadro's constant to get the number ions.

No of ions= No of moles × L

L is the Avogadro's number.

3 0
3 years ago
Read 2 more answers
If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?
Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L  - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

new volume -  125 mL + 25 mL = 150 mL

therefore new molarity is

c = 0.01875 mol / 0.150 L  = 0.125 M

answer is 0.125 M

7 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
Dependent variable? ( what did we measore)?
lana66690 [7]
The thing u change.
6 0
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I need help with this question please help me
bulgar [2K]

8xc take away 8xExplanation:

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