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Whitepunk [10]
3 years ago
13

Average atomic mass work

Chemistry
1 answer:
babymother [125]3 years ago
7 0
Take 234/100 x 0.01
then 235/100 x 0.71
then 238/100 x 99.28
you divide by 100 because the abundance is given in percentage over 100
then add up all the values, you should get an answer of 237.97 rounded up is 238.
Final answer : 238
hope this helped :)

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What is the term for the process that occurs when an atom of an element spontaneously breaks apart to create an atom of a differ
quester [9]
Atoms of elements below atomic number 20 do not split because of stable nucleus.
Stability of nucleus is determined by neutron/proton ratio.
For light stable nuclides, number of protons ans neutrons are same. For heavy nuclides, number of neutrons is more than the number of protons.

Here the type of decay is alpha decay because number of nucleons(protons and neutrons) changes.
Hence option d is correct.
8 0
3 years ago
A(n) _________is a material that contains only one type of atom.
joja [24]

Answer:

element

Explanation:

Elements are pure substances with only one kind of atom

7 0
4 years ago
How many moles does 21. 7 L of gas at STP represent?
Oduvanchick [21]

Answer:

.96875 moles

Explanation:

22.4 liters per mole at STP for a gas

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3 0
2 years ago
Decide which of the following statements are True and which are False about equilibrium systems:A large value of K means the equ
ivanzaharov [21]

Answer:

a. True

b. False

c. True

d.  False

e. False

Explanation:

A. (true) The equilibrium constant K is defined as

\frac{Products}{reagents}

In any case  

aA +Bb ⇌ Cd +dD

where K is:

K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

A large value on K means that the concentration of products is bigger than the concentrations of reagents, so the forward reaction is favored, and the equilibrium lies to the right.

B. (False) When we work with gases, we use partial pressure to make calculations in the equilibrium, so we estimate Kp as:

Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}

Using the ideal gas law, we can get a relationship between K and Kp  

Pv=nRT where P=\frac{n}{v}*RT we know that \frac{n}{v} is the molar concentration. When we replace P in the expression for Kp we get:

Kp= \frac{[C]^{c}*(RT)^{c}[D]^{d}*(RT)^{d}}{[A]^{a}*(RT)^{a}[B]^{b}*(RT)^{b}}

Reorganizing the equation:

Kp= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}*\frac{(RT)^{c+d}}{(RT)^{a+b}}

We can see K in the expression  

Kp= K*(RT)^{c+d-a-b}

Delta n = c+d-a-b

Kp= K*(RT)^{delta n}

For the reaction  

H_{2}(g) + F_{2}(g)-- equilibrium---2HF(g)

Delta n = 2-1-1=0

Kp= K*(RT)^{0}

So Kp=K in this case.

C. (true) The value of K just depends on the temperature that’s why changing the among of products won’t have any effect on its value.  

D. (false) as we can see this reaction involve a heterogeneous system with solids and gases. For convention the concentration for solids and liquids can be considered constant during the reaction that’s why they’re not include in the calculation for the equilibrium constant. Taking this into account the expression for the equilibrium for this reaction is:

CaCO_{3}(s)---equilibrium----CaO(s) + CO_{2}(g)

K= [CO_{2}]

So we can see that [CaCO_{3}] is not include in the expression.  

E. (False) The equilibrium is defined as the point where the rate of the forward reaction is the same to the rate of the reverse reaction. The value of K is telling you which reaction is favored but the rate of both reactions is the same in this point. (see picture)  

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Answer:

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