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lara31 [8.8K]
2 years ago
5

What is the purpose of the dts other auths and pre audits screen?

Chemistry
1 answer:
ICE Princess25 [194]2 years ago
6 0

Answer:

It allows the traveler to justify questionable expense items.

hope that helped <3

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A geologist is studying the shore along a river. She finds a pile of rocks at the base of a riverbank. These broken rock pieces
Dmitry_Shevchenko [17]

The scientist who studies the earth constituent like solids, gas and liquid is called a geologist. Weathering and erosion are the correct blanks.

<h3>What are weathering and erosion?</h3>

The process of the deterioration of any solid particles like minerals, soils and rocks is called weathering. It occurs due to biochemical substances, atmospheric gases and water pressure.

The process of the transportation of the rocks and mineral particles away from their original position is called erosion it is due to forceful air and water.

The rocks moved from their original place to some another due to weathering and this process of their transportation is called erosion.

Therefore, <u>weathering </u>and <u>erosion </u>are the correct blanks.

Learn more about weathering and erosion here:

brainly.com/question/11712593

7 0
2 years ago
What amount of water is formed when 20 ml of 0.80 m hcl and 30 ml of 0.40 m naoh are mixed?
Yuri [45]

Answer: The amount of water formed is 12 moles

Explanation: Please see the attachments below

7 0
3 years ago
You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of
SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

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

Or,

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

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

Moles=1.60 \times {300.0\times 10^{-3}}\ moles

<u>Moles of potassium iodide = 0.48 moles </u>

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles=1.20\times {285\times 10^{-3}}\ moles

<u>Moles of lead(II) nitrate  = 0.342 moles </u>

According to the given reaction:

2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

<u>Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.</u>

Also, consumed lead(II) nitrate = 0.24 moles  (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.102/0.585 M = 1.174 M</u>

<u>Statement A is correct.</u>

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

<u>Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g </u>

<u>Statement B is correct.</u>

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.48/0.585 M = 0.821 M</u>

<u>Statement C is correct.</u>

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate  produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate  produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

<u>So, Concentration = 0.684/0.585 M = 1.169 M</u>

<u>Statement D is incorrect.</u>

4 0
2 years ago
The inhibition caused by the final end product of a reaction is called
Troyanec [42]

Answer:

Non competitive inhibition

Explanation:

Hello,

During enzymatic catalysis, the active sites could be occupied by the very same products' molecules turning out into an inhibition (the reaction starts to slow down since to active places are available for the reagents to react). Nonetheless this inhibition is not competitive as long as the product does not react due to the active sites it is occupying.

Best regards.

5 0
2 years ago
Which is associated with a fission reaction? *
podryga [215]

Answer: Radioactive waste

Explanation:

The nuclear fission reaction consists of heavy atomic particles or heavy nucleus, like plutonium and uranium and in radioactive heavy metals. In the fission reaction the nucleus get split into equal masses of particles. This process is associated with release of large amount of energy. The fission of radioactive waste can cause deadly mutations in living beings.      

6 0
3 years ago
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