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mezya [45]
2 years ago
14

List the sequence of events that occurred in the area represented by the profile above.

Chemistry
1 answer:
Fittoniya [83]2 years ago
7 0

1. Igneous intrusion Y formed.

2. The shale layer was deposited.

3. Igneous intrusion X formed.

4. The limestone layer was deposited.

5. The mudstone layer was deposited.

B 1. The mudstone layer was deposited.

2. The limestone layer was deposited.

3. The shale layer was deposited.

4. Igneous intrusion Y was forced onto the mudstone, limestone, and shale.

5. Igneous intrusion X was forced onto the mudstone, limestone, and shale.

C 1. Igneous intrusion X formed.

2. Igneous intrusion Y formed.

3. The shale layer was deposited.  

4. The limestone layer was deposited.

5. The mudstone layer was deposited.

D 1. The layer of mudstone was deposited.

2. The layer of limestone was deposited.

3. Igneous intrusion Y was forced onto the mudstone and limestone.

4. The shale layer was deposited.

5. Igneous intrusion X was forced over the mudstone, limestone, and shale

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How many grams of fluorine are contained in 8 molecules of boron trifluoride?
Lelu [443]
<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

                 Number of Molecules  =  8

                 M.Mass of BF₃ =  67.82 g.mol⁻¹

                 Mass of Fluorine atoms  =  ?

Step 1: Calculate Moles of BF₃

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

4 0
3 years ago
Read 2 more answers
The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?
AlladinOne [14]

Answer:

64.52 mg.

Explanation:

The following data were obtained from the question:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Final amount (N) =.?

Next, we shall determine the rate constant (K).

This is illustrated below:

Half life (t½) = 1590 years

Rate/decay constant (K) =?

K = 0.693 / t½

K = 0.693/1590

K = 4.36×10¯⁴ / year.

Finally, we shall determine the amount that will remain after 1000 years as follow:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Rate constant = 4.36×10¯⁴ / year.

Final amount (N) =.?

Log (N₀/N) = kt/2.3

Log (100/N) = 4.36×10¯⁴ × 1000/2.3

Log (100/N) = 0.436/2.3

Log (100/N) = 0.1896

Take the antilog

100/N = antilog (0.1896)

100/N = 1.55

Cross multiply

N x 1.55 = 100

Divide both side by 1.55

N = 100/1.55

N = 64.52 mg

Therefore, the amount that remained after 1000 years is 64.52 mg

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Answer:

division of atoms are neither created nor destroyed therefore atoms can only be rearranged

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