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Rina8888 [55]
2 years ago
8

What is caused by temperature differences in the mantle resulting in plate movement?

Chemistry
1 answer:
AlekseyPX2 years ago
7 0

Answer:

Convection Currents

Explanation:

You might be interested in
Assume you find four bottles in an empty laboratory, each containing a liquid. The labels that were on these bottles have fallen
fenix001 [56]

Answer:

No, I can not identify the contents of each bottle using solubility and polarity (with H2O) information

Explanation:

While it is true that polar substances dissolve in water and nonpolar substances do not dissolve in water, the task here is to specifically identify the contents of each of the bottles.

Solubility in water can not tell us exactly what liquid is which substance. For instance, trans-1,2-dichloroethylene, cis-1,2-dichloroethylene and cyclooctane are all insoluble in water. The fact that they do not dissolve in water does not tell us which liquid is which compound.

Even though acetic acid is miscible with water, it is not a conclusive prove that the liquid is acetic acid since other polar organic compounds are also miscible in water.

It is only by determining the boiling point of each substance that I can conclusively identify the contents of each bottle since boiling point is an intrinsic property of substances.

6 0
2 years ago
Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
5 0
3 years ago
Read 2 more answers
Which type of stoichiometry problems does not require the use of molar mass?
Arlecino [84]

Answer:

Explanation:

It is volume-volume problems that does not require the use of molar mass.

7 0
3 years ago
The following data was collected when a reaction was performed experimentally in the laboratory.
Lera25 [3.4K]

Answer:

Mass = 279.23 g

Explanation:

Given data:

Number of moles of Fe₂O₃ = 3 mol

Number of moles of Al = 5 mol

Maximum amount of iron produced by reaction = ?

Solution:

Chemical equation:

Fe₂O₃ + 2Al    →     Al₂O₃  +  2Fe

Now we will compare the moles of iron with Al and iron oxide.

                          Fe₂O₃     :       Fe

                             1           :        2

                             3          :       2×3 = 6 mol

                            Al          :          Fe

                              2         :           2

                               5        :           5 mol

The number of moles of iron produced by Al are less so Al is limiting reacting and it will limit the amount of iron so maximum number of iron produced are 5 moles.

Mass of iron:

Mass = number of moles × molar mass

Mass =   5 mol  ×  55.845 g/mol

Mass = 279.23 g

 

3 0
2 years ago
2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.
AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

7 0
3 years ago
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