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Oduvanchick [21]
3 years ago
7

The two isotopes of potassium with significant abundance in nature are 39K asotopic mass 38.9637 amu, 932.58%) and 41K Osotopic m

ass 40.9618 amu, 6.730%). Fluorine has only one naturally occurring isotope, 19F (isotopic mass 18.9984 amu). Calculate the formula mass of potassium fluoride.
Chemistry
1 answer:
Komok [63]3 years ago
4 0

Answer:

\boxed{\text{58.0919 u}}

Explanation:

1. Atomic mass of K

To get the atomic mass of K, you calculate the weighted average of the isotopic masses.

That is, you multiply the atomic mass of each isotope by a number representing its relative importance (e.g., its percent of the total).

Set up a table for easy calculation:

\begin{array}{ccrcr}\textbf{Atom} & \textbf{Mass/u} &\textbf{Percent} & \textbf{Calculation}& \textbf{Result/u}\\^{39}\text{K}& 38.9637 & 93.258 & 38.9637 \times 0.932 58 & 36.3368\\^{41}\text{K}& 40.9618 & 6.730 & 40.9618 \times 0.06730& 2.7567\\& & \text{TOTAL } = &\textbf{39.0935}\\\end{array}

2. Formula mass of KF

\begin{array}{rcl}\text{K} & = & \text{39.0935 u}\\\text{F} & = & \text{18.9984 u}\\\text{Total} & = & \textbf{58.0919 u}\\\end{array}\\\text{The formula mass using only the two isotopes of K is $\boxed{\textbf{58.0919 u}}$}

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What is Haber's process? Write it with chemical reaction.​
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7 0
3 years ago
Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Which ion would NOT be present in
Allisa [31]

Answer:

The question is incomplete and confusing.

  • In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.

  • In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.

See below the details.

Explanation:

Which compound will not form ions?

<u />

<u>1. Write the balanced molecular equation:</u>

  • Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

<u />

<u>2. Write the ionizations for the ionic aqueous compounds:</u>

<u />

  • Pb(NO₃)₂(aq) →  Pb⁺²(aq) + 2NO₃⁻(aq)

  • 2KI(aq) → 2K⁺(aq) + 2I⁻(aq)

  • 2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)

<u />

<u>3. Write the complete ionic equation:</u>

Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) +  2K⁺(aq) + 2NO₃⁻(aq)

Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.

All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.

It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.

3 0
3 years ago
How many moles of H2SO4 are present in 3.63 liters of a 0.954 M solution?
azamat
For this problem, we will use the formula setup for Molarity. 
The formula is M= moles
                               ---------
                                liters

So, lets plug in the numbers we have. We know M= 0.954, and L (liters)= 3.63. Lets plug that into our equation.
                         
                         x
         0.954= -----
                     3.63

As we can see, our moles are unknown.
So, lets multiply 0.954 times 3.63 to get our moles. 
0.954 x 3.63 equals 3.46302.

We know this is our correct answer, because when we divide 3.46302 and 3.63, we get 0.954.

Why?
Because when we divide the moles by liters, we get the molarity.

I hope I helped!!!!!

Here are some common units to know if my units confused you.
M= Molarity, which is what we find by dividing moles by liters
mol= Moles
L= Liters

5 0
4 years ago
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