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3 years ago

Explain why the sun will not explode at any point in its life cycle

Chemistry

2 answers:

vodka [1.7K]3 years ago

7 0

Answer:

i no now dethankyou

Explanation:

Setler79 [48]3 years ago

3 0

The sun will explode in its life cycle but maybe not now but probably in another million years and for right now it won’t since the gasses and rocks it has aren’t as compacted together

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URGENT!!!! Please respond in less than 5 mins hurry!!!!?!!?!!??!?

AleksAgata [21]

Answer:

1. A

2. B

Explanation:

1. Most of the answers don't make that much sense, so using process of elimination I resulted in this answer. I recommend you also try to check with others.

2. This question also was pretty confusing but when using process of elimination I resulted in either A or B. To me B makes more sense.

5 0

3 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

3 years ago

When the equation ZnS+O ZnO+SO isbalanced the coefficientof O is 1, 5, 3,or2

Flauer [41]

ZnS + O --> ZnO + SO

Okay so first you have to count up the number of elements on each side of the equation. Your objective is to have the same number of each element on both sides.

Left Side:
Zn - 1
S - 1
O - 1

Right Side:
Zn - 1
S - 1
O - 2

Since there are two oxygens on the right side, you have to add a coefficient of 2 to the oxygen on the left side. The coefficient tells us that that element or molecule is being multiplied by the value of coefficient. Since we're adding a coefficient of 2 to the oxygen on the left side, there are now 2 oxygens on that side. Because that is the same amount of oxygen as on the right, the equation is now balanced.

Your final equation should look like this: ZnS + 2O --> ZnO + SO

In conclusion, the answer is 2.

3 0

3 years ago

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An experimenter begins with two equal balls of clay and then changes the shape of one the experiment is most likely testing for

zhenek [66]

I believe the answer is conservation of matter. I hope this helps!

7 0

4 years ago

In your own words, explain how you would make an electromagnet.

notsponge [240]

Electromagnets can be created by wrapping a wire around an iron nail and running current through the wire.

4 0

3 years ago