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PolarNik [594]
2 years ago
7

DUDE SOMEONE HELP ill love u forever

Chemistry
1 answer:
dybincka [34]2 years ago
4 0

Answer:

uwu good

Explanation:

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Four modes of sediment transport are:
netineya [11]
Here is 5


Dissolved Load - elements dissolved in solution

Suspended Load - very fine grained sediment such as clay and silt carried in suspension. The size grains that can be carried in suspension are dependent on the current velocity

Wash Load - a subset of the suspension load, extremely small particles (clay) that will remain in suspension independent of turbulence in the river

Saltation Load - particles that are temporarily carried in suspension but move by bouncing along the bottom
 
<span>Bed Load - sediment that moves by rolling or sliding along the bottom. These are generally the coarser grained sediments such as sand and gravel.</span>
5 0
2 years ago
Explain the difference between a base and an alkali.
Nataly_w [17]

Answer:

a base is something that reacts with an acid to form water and salt , an alkai is any base that is soluble in water

5 0
2 years ago
At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con
djyliett [7]
0.114 mol/l  
The equilibrium equation will be: 
Kc = ([Br2][Cl2])/[BrCl]^2  
The square factor for BrCl is due to the 2 coefficient on that side of the equation.  
Now solve for BrCl, substitute the known values and calculate. 
Kc = ([Br2][Cl2])/[BrCl]^2 
[BrCl]^2 * Kc = ([Br2][Cl2]) 
[BrCl]^2 = ([Br2][Cl2])/Kc 
[BrCl] = sqrt(([Br2][Cl2])/Kc)  
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142) 
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142) 
[BrCl] = sqrt(0.013021127 mol^2/l^2) 
[BrCl] = 0.114110152 mol/l  
Rounding to 3 significant figures gives 0.114 mol/l
4 0
3 years ago
Which element is presented in all organic compounds
DerKrebs [107]
Carbon and hydrogen
3 0
3 years ago
A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex
Anestetic [448]

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

8 0
3 years ago
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