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pogonyaev
3 years ago
10

What holds all the planets and objects in the solar system in place?

Chemistry
1 answer:
uysha [10]3 years ago
5 0

Answer:

Gravity

Explanation:

i dont rlly know where it comes from...

You might be interested in
Why do you think the older scientist disregarded the younger scientist's prediction?
Svetach [21]

Answer:

The older scientist were ignorant

Explanation:

they thought that they were better than the younger scientist. to put it in simply they were ignorant to think younger is worse

5 0
3 years ago
Read 2 more answers
The work of scientists usually begins with
VladimirAG [237]
A. Testing a hypothesis

This leads to the other points
6 0
4 years ago
Which one of the following is an oxidation-reduction reaction?
kupik [55]

Answer:

CH4 + 2 O2 --> CO2 + 2 H2O

Explanation:

CH4 + 2 O2 --> CO2 + 2 H2O is the only reaction where an element (oxygen) undergoes a change in oxidation state. In this reaction oxygen changes disproportionately to O⁻². That is ...

O₂ → CO₂ + 4e⁻ ==> oxidation

<u>O₂ + 4e⁻  →  H₂O ==> reduction </u>

2O₂ + 4e⁻  →  CO₂ + H₂O + 4e⁻  ==> Net oxidation-reduction

=>  4e⁻ gained by one mole O₂ in formation of CO₂ = 4e⁻ lost by the other mole O₂ in forming H₂O.

Then...

Including CH₄ (whose elements do not undergo changes in oxidation states) requires doubling reaction to balance by mass thus giving ...

2CH₄ + 2O₂ + 8e⁻  →  2CO₂ + 2H₂O + 8e⁻

Cancelling 8 reduction electrons on left with 8 oxidation electrons on right gives...

2CH₄ + 2O₂  →  2CO₂ + 2H₂O

7 0
4 years ago
Read 2 more answers
Determine the location of the last significant place value by placing a bar over the digit.
Vlad1618 [11]

Answer:

8,040  

0.0300  

699.5  

2.000 x 102

0.90100  

90, 100  

4.7 x 10-8  

10,800,000.0  

3.01 x 1021

0.000410

Explanation:

First remember the following rules of determining the last significant place value :

1. The digits from 1-9 are all significant and zeros between significant digits are also significant.

2.  The trailing or ending zeroes are significant only in case of a decimal number otherwise they are ignored. However starting zeroes of such a number are not significant.

Now observing above rules, lets determine the location of the last significant place value of each given example. I am determining the location by turning the last significant place to bold.

1) 8,040

8,040

Location of the last significant place value is 3 and bar is over last significant digit that is 4. Number is not decimal so ending zero is ignored. Every non zero digit is a significant.

2)  0.0300

0.0300

Location is 3 and bar is over 0. Number has a decimal point so ending zero is not ignored but starting zeroes are ignored.

3) 699.5

699.5

Location is 4 and bar is over 5.

4) 2.000 x 10²

2.000 x 10²

Location is 4 and bar is over 0. This is because the number is decimal so trailing zeroes cannot be ignored. Also if we convert this number it becomes:

200.0 so last significant digit is 0 and location of last significant digit is 4.

5) 0.90100

0.90100

Location is 5 and bar is over 0. This is because in a number with decimal point starting zeroes are ignored but trailing zeroes after decimal point are not ignored. So we count from 9 and last significant digit is 0.

6) 90, 100

90, 100

Location is 3 and bar is over 1. This is because it is not a number with decimal point. So the trailing zeroes are ignored. The count starts from 9 and last significant is 1.

7) 4.7 x 10⁻⁸

4.7 x 10⁻⁸

Location is 2 and bar is over 7. This is because the starting zeroes in a number with a decimal point are ignored. So the first digit considered is 4 and last significant digit is 7. If we expand this number:

4.7 x 10⁻⁸ =    0.000000047 = 0.000000047

Here the starting zeroes are ignored because there is a decimal point in the number.

8) 10,800,000.0

10,800,000.0

Location is 9 and bar is over 0.  Number has a decimal point so ending zero is not ignored and last significant figure is 0.

However if the number is like:

10,800,000. Then location would be 8 and bar is over 0.

9) 3.01 x 10²¹

3.01 x 10²¹

Location is 3 and bar is over 1. Lets expand this number first

3.01 x 10²¹ = 3.01 x 1000000000000000000000

                  = 3010000000000000000000

So this is the number:

3010000000000000000000

Since this is number does not have a decimal point so the trailing zeroes are ignored. Hence the count starts from 3 and the last significant figure is 1

10) 0.000410

0.000410

Location is 3 and bar is over 0. This is because the number has a decimal point so the ending zero is not ignored but the starting zeroes are ignored according to the rules given above. Hence the first significant figure is 4 and last significant figure is 0.

6 0
3 years ago
4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?
melomori [17]

Answer: 9.08 L

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles

4Al+3O_2\rightarrow 2Al_2O_3

According to stoichiometry :

4 moles of Al require  = 3 moles of O_2

Thus 0.54 moles of Al will require=\frac{3}{4}\times 0.54=0.405moles  of O_2

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L

Thus 9.08 L of O_2 at STP would be required

6 0
3 years ago
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