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3 years ago

What was noticed about the shape of coastlines by sailors as early as the sixteenth century?

Chemistry

2 answers:

Wewaii [24]3 years ago

8 0

Answer:

The East coastline of South America and West coastline of Africa looked like they fit together

Explanation:

alukav5142 [94]3 years ago

7 0

One major significant geographical feature that sailors during the Age of Exploration noticed were that the Eastern coastline of South America, fits like a puzzle to the Western coast of Africa. Furthermore, this observation firmly supported the Continental Drift theory of the origin of Earth's continents.

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1.20×10−8s to nanoseconds

AveGali [126]

Answer:

There are 12 nanoseconds in $1.2\times 10^{-8}\ s$ .

Explanation:

We need to convert $1.2\times 10^{-8}\ s$ to nanoseconds.

We know that,

$1\ s=10^9\ ns$

Now using unitary method to solve it such that,

$1.2\times 10^{-8}\ s=1.2\times 10^{-8}\ \times 10^9\\\\=1.2\times 10\\\\=12\ ns$

So, there are 12 nanoseconds in $1.2\times 10^{-8}\ s$ .

7 0

3 years ago

What type of radioactive decay will the isotopes 13B and 188Au most likely undergo?

scoundrel [369]

Answer:

b. Beta emission, beta emission

Explanation:

A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).

Now let us look at the N/P ratio of each atom;

For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6

For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4

For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.

For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.

8 0

3 years ago

What is the correct sequence of coefficients when this equation is balanced? ___cs2(l) + ___o2(g) → ___co2(g) + ___so2(g)?

iris [78.8K]

Answer: the sequence is 1, 3, 1, 2.

Explanation:

1) Given equation: CS₂(l) + O₂(g) → CO₂(g) + SO₂(g)

2) To balance:

i) initially C is balanced: 1 atom in the left side, and 1 atom in the right sides

ii) Add coefficient 2 in front of SO₂ to have 2 atoms in each side.

That leads to: CS₂(l) + O₂(g) → CO₂(g) + 2SO₂(g)

iii) Count the O atoms; there are 2 atoms in the left, and 6 atoms in the right side. So, add a coefficient 3 in front of O₂ (in the left) and you get:

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)

4) You can verigy that now the equation is balanced:

C: 1 in the left, and 1 in the right

S: 2 in the left, and 2 in the right

O: 6 in the left, and 6 in the right.

5) So, the sequence is 1, 3, 1, 2.

Remember, when the none coefficient is shown it is because it is 1.

4 0

3 years ago

Read 2 more answers

What is the main function of the part labeled Y in the model?

Tanya [424]

Answer

Explanation

the answer is D

8 0

3 years ago

Read 2 more answers

10g of an oxide of copper contains 8.882g of copper. what is empirical formula of the copper oxide

-BARSIC- [3]

Answer:

Mass of copper oxide: 10g

Mass of copper: 8.882g

Mass of oxygen: 1.118g

Cu O

8.882 1.118

Divide by $A_{r}$ 8.882÷63.5 1.118÷16

0.14 0.07

Divide by the lowest number 0.14/0.07 0.07/0.07

2 1

Check the ratio now, Cu has 2 atoms and O has 1 atom .

The formula is $Cu_{2} O$

4 0

3 years ago