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tino4ka555 [31]
3 years ago
14

help plz Which list is in order from smaller to larger sediment particles? A. sand, silt, clay B. clay, silt, sand C. silt, sand

, clay D. silt, clay, sand
Chemistry
1 answer:
MariettaO [177]3 years ago
5 0

Answer : The correct option is B (clay, slit, sand).

Explanation :

Soil particles are of three types : Sand, Slit and clay. Most of the soils particles are made up of a combination of sand, slit and clay.

Particle size range of the sand = 2.00 -0.05 mm

Particle size range of the slit = 0.05 -0.002 mm

Particle size range of the sand = less than 0.002 mm

Sand is the largest soil particle, clay is the smallest soil particle and slit particle is present in between the sand & clay particle.

Therefore, the order from smaller to larger sediment particles is

Clay > Slit > Sand

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What will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?
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THE VOLUME OF THE NITROGEN GAS AT 2.5  MOLES , 1.75 ATM AND 475 K IS 55.64 L

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Using the ideal gas equation

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6 0
3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

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