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ASHA 777 [7]
3 years ago
6

Sea floor spreading occurs at a mid-ocean ridge true or false

Chemistry
1 answer:
Elenna [48]3 years ago
7 0

Answer:

False

Explanation:

sea floor spreading is not consistent at all mid ocean ridges.

You might be interested in
The three beakers shown below contain solutions of [cof6]3–, [co(nh3)6]3+, and [co(cn)6]3–. based on the colors of the three sol
vitfil [10]
[Co(CN)₆]³⁻ → Yellow
[Co(NH₃)₆]³⁺ → Orange
[CoF₆]³⁻ → Blue
Explanation:
- All the given compounds have octahedral geometry but the ligand in each are different with the same metal ion.

- Ligands strength order:     CN⁻ > NH₃ > F⁻ 

- The ligand CN will act as a strong field ligand so that the splitting is maximum when compared to NH₃ and F⁻

- If the splitting is more, the energy required for transition is more, and the wavelength is inversely proportional to energy.

- So CN complex will absorb at lower wavelength (yellow color)
3 0
3 years ago
how much hydrogen gas is necessary to exert a pressure of 1.4 ATM at 430k if occupying a volume of 15.1 l
lora16 [44]

Answer:

5

Explanation:

7 0
3 years ago
Calculate the number of formula units in 2.50 mol NaNO
marta [7]

The number of formula units in 2.50 mol of the compound is  15.1 * 10^23.

The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.

6.02 * 10^23 formula units of the compound are contained in 1 mole

x formula units are contained in 2.5 moles of the compound

x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole

x = 15.1 * 10^23 formula units of the compound.

Learn more; brainly.com/question/9743981

3 0
2 years ago
A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume
jenyasd209 [6]
Volume of Argon V1 = 5.0 L  
Pressure of Argon P1 = 2 atm 
Final temperature T2 = 30 C = 30 + 273 = 303 K 
Volume at final temperature V2= 6 L 
Pressure at final temperature P2 = 8 atm  
We know that (P1 x V1) / T1 = (P2 x V2) / T2  
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48 
Initial Temperature T1 = 3030 / 48 = 63.12 
Initial Temperature = -209. 8 C
4 0
3 years ago
Determine the [h3o+] of a 0.210 m solution of formic acid.
Nataly [62]
When the Pka for formic acid = 3.77
and Pka = -㏒ Ka 
   3.77 = -㏒ Ka
∴Ka = 1.7x10^-4 

when Ka = [H+][HCOO-}/[HCOOH]

when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2      by solving this equation for X

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
       = -㏒ 0.0059
       = 2.23 

3 0
3 years ago
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