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makkiz [27]
2 years ago
13

A vial contains radioactive iodine-131 with an activity of 2.0 mCi/mL. If a thyroid test requires 2.9 mCi in an "atomic cocktail

," how many milliliters are used to prepare the iodine-131 solution?
Chemistry
1 answer:
stira [4]2 years ago
7 0

Answer:

1.45 milliliters

Explanation:

The given question states that a vial contains radioactive iodine-131 with has an activity of 2.0 mCi/mL. Therefore, if a thyroid test requires 2.9 mCi in an "atomic cocktail", the number of milliliters used to prepare the iodine-131 solution is estimated below:

Using simple mathematics

For an activity of 2.0 mCi/mL, it means that I ml can be used to prepare 2.0 mCi of iodine-131, thus, for 2.9mCi, the number of milliliters required will be:

2.9 mCi/(2.0mCi/mL) = 1.45 mL.

Therefore, 1.45 milliliters are required to prepare 2.9 mCi of iodine-131 solution.

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An isotope undergoes radioactive decay by emitting radiation that has no mass. What other characteristic does the radiation have
Advocard [28]

Answer : Option D) No charge

Explanation : An isotope undergoes radioactive decay by emitting radiation that has no mass. The radiation will not have any charge as it does not has any mass it will not emit a radiative charge.

It is known that there are some unstable radioactive isotopes which has no mass and the radiation thus has no charge in it.

7 0
2 years ago
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Determine the y component of reaction at a using scalar notation. express your answer to three significant figures and include t
blagie [28]
In order to compute the y-component of a vector, we simply use the formula:

Fy = F*sin(∅)
Where ∅ is the angle of the vector measured from the positive x-axis and F is the magnitude of the vector.

Similarly, the x-component is calculated by substituting sin(∅) with cos(∅)
6 0
2 years ago
Bonds between two atoms that are equally electronegative are _____. bonds between two atoms that are equally electronegative are
Anna [14]
Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.

Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."
4 0
2 years ago
Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 87ml+87ml=174ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 174 g

T_{final} = final temperature of water = 317.4 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=174g\times 4.18J/g^oC\times (317.4-298)K

q=14110.008J=14.11KJ

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole

Therefore, the enthalpy of neutralization is, -101.37 KJ

3 0
2 years ago
Be sure to answer all parts. liquid ammonia autoionizes like water: 2nh3(l) → nh4+(am) + nh2−(am) where (am) represents solvatio
Genrish500 [490]
Answer is: concentration of ammonium ions are 7,14·10⁻¹⁴ M.
Chemical reaction: 2NH₃(l) → NH₄⁺(am) + NH₂⁻(am).
Kam = 5,1·10⁻²⁷.
[NH₄⁺] · [NH₂⁻] = x; equilibrium concentration of cations and anions.
Kam = [NH₄⁺] · [NH₂⁻].
Kam = x².
x = [NH₄⁺] = √5,1·10⁻²⁷.
[NH₄⁺] = 7,14·10⁻¹⁴ M.

5 0
2 years ago
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