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Zinaida [17]
3 years ago
9

In order for a hypothesis to become a theory, scientists MUST _____. A. do many experiments B. find a well-defined question C. m

ake their question testable D. throw out their earlier experimental results
Chemistry
2 answers:
Dovator [93]3 years ago
8 0

the answer is A do many experiments.

Tamiku [17]3 years ago
7 0
A. If there is enough evidence to support a hypothesis it becomes a theory. In order to get evidence a scientist would have to perform many experiments.
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Beth pushes a box with a force of 2 N to the left. Dave pushes the same box from the opposite side with a force of 8 N to the ri
AleksAgata [21]

Answer:

is this based on the newtons law and balnce force

Explanation:

5 0
3 years ago
What form of waste cannot be recycled to make new products?
skelet666 [1.2K]
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4 0
3 years ago
Read 2 more answers
Name each of the following molecular binary compounds:
Tanya [424]

Answer:

Explanation:

From the given information; we are to assign the correct naming for the molecular binary compounds.

01. CaCl2 → Calcium CHloride

02. HI (g) → Hydrogen mono iodide gas

03. HI (aq) →  Hydroiodic acid

04. BH3 → Borontrihydride

05. Cl2O6 → Dichlorine hexoxide

06. ClF → Chlorine monoflouride

07. P2Cl4 → Phosphorus tetrachloride

08. I2O5 → Diiodine pentaoxide

09. BrF5 → Bromine pentaflouride

10. HBr(aq) → Hydrobromic acid

11. IF7 → Iodine heptaflouride

12. NF3 → nitrogen triflouride

13. H2Se(aq) → hydroselenic acid

14. BrCl → Bromine monochloride

15. H2Se(g) → Hydrogen Selenide

16. SnO2 → Tindeoxide

17. HBr(g) →  Hydrogenbromide gas

18. AsF3 → Arsenic triflouride

19. N2O3 → Dinitrogentrioxide

20. IF4 → Iodine pentafluoride

01. hydroiodic acid  → HI (aq)

02. hydrogen monoiodide gas → HI (g)

03. diiodine tetraoxide → I2O4

04. bromine monofluoride → BrF

05. silicon tetrahydride  → SiH4

06.  tetraphosphorus hexaoxide → P4O6

07. disulfur monoxide → S2O

08. carbon monooxide → CO

09. hydrogen monochloride gas → Hcl

10. tetraphosphorus decaoxide → P4O10

11. dibromine monoxide → Br2O

12. dinitrogen tetrafluoride → N2F4

13. disilicon hexahydride → SiH6

14. tetraarsenic hexaoxide → As4O6

15. hydrochloric acid → HCl (aq)

16.  arsenic trihydride → AsH3

17. iodine heptafluoride → IF7

18. bromine dioxide → BrO2

19. disulfur decafluoride → S2F10

20. dichlorine heptaoxide → Cl2O7

4 0
3 years ago
Which of the following represent impossible combinations of n and I? (a) 1p, (b) 4s, (c) 5f, (d) 2d 6.61
Eduardwww [97]

Answer:

  • <u>Options (a) and (d) : (a) 1p and (d) 2d represent impossible combinations of n and l.</u>

Explanation:

n refers to the principal quantum number, and l refers to the angular momentum or Azimuthal quantum number.

Principal quantum number (n) is used to indicate the main energy level of the electron. It may take whole numbers: 1, 2, 3, 4, 5, 6, 7, ...

Angular momentum or Azimuthal quantum number refers to the kind (shape) of the orbital. It can take numbers from 0 to n - 1.

So, if n = 1, l can only be 0; if n = 2, l can be either 0 or 1; if n = 2, l can be either 0, 1 or 2.

On the other hand, the shape of the orbitals is also representd by a letter. then there is a unique relation between the letter that represents the orbital and the angular quantum number which is:

letter       l number

s               0

p               1

d               2

f                3

The previous information is summarized in the next table:

n    possible l numbers          

1          0                                          

2         0, 1                                        

3         0, 1, 2                                    

4         0, 1, 2, 3                              

5        0, 1, 2, 3, 4

6        0, 1, 2, 3, 4, 5

7        0, 1, 2, 3, 4, 5, 6

As per the choices given in the question you have:

<u>(a) 1 p</u> is not possible because when n = 1 the only l number is 0 and it is an s orbital, but p ⇒ l = 1. Thus, this is a correct choice.

<u>(b) 4s</u> is possible since n = 4 permits l to be 0, 1, 2, and 3.

<u>(c) 5f </u>is possible since n = 5 permits l to be 0, 1, 2, 3, and 4.

<u>(d) 2d</u> is impossible since n = 2 permits l to be 0, and 1, but d ⇒ l = 2. Thus, this is other right choice.

4 0
3 years ago
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