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emmasim [6.3K]
3 years ago
7

Will give brainliest

Chemistry
1 answer:
slamgirl [31]3 years ago
3 0
I will help you with answering this question.
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Answer:

1. 2,3,2

2. 4,5,2

3. 3,1,1

Explanation:

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How many ligaments in e kne
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Enter the appropriate symbol for an isotope of calcium-42 corresponding to the isotope notation AZX
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<span>40 20Ca this is the appropriate symbol for an isotope of calcium-42 corresponding to the isotope notation AZX</span><span />
3 0
3 years ago
Read 2 more answers
Potassium chlorate is a white, crystalline solid. It can be broken down into potassium chloride (a salt)
lions [1.4K]

Answer:

Notice that the number of atoms of

K

and

Cl

are the same on both sides, but the numbers of

O

atoms are not. There are 3

O

atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6

O

atoms on both sides.

2KClO

3

(

s

)

+ heat

→

KCl(s)

+

3O

2

(

g

)

Now the

K

and

Cl

atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of

KCl

.

2KClO

3

(

s

)

+ heat

→

2KCl(s)

+

3O

2

(

g

)

The equation is now balanced with 2

K

atoms,

7 0
3 years ago
Read 2 more answers
What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0
3 years ago
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