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Mrrafil [7]
2 years ago
12

Which two parts of sedimentary rock formation include the breakdown and carrying away of existing rock?

Chemistry
2 answers:
mamaluj [8]2 years ago
7 0
Erosion and weathering
Rzqust [24]2 years ago
3 0

erosion and weathering.

erosion as in breaking it down, weathering is also a cause of erosion, but in the case of rain, it would carry away the remaining sediments.

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Are spotted cats cuter than solid-colored cats?

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3 years ago
Which of the following are considered pure substances?
Helen [10]
B.Elements

Explanation: they cannot be separated
8 0
3 years ago
What is the concentration, in mass percent (m/m), of a solution prepared from 50.0 g nacl and 150.0 g of water?
Nataly [62]
Mass of solute ( m1 ) = 50.0 g

mass of solvent ( m2 ) = 150.0 g

Therefore:

m/m = ( m1 / m1 + m2 )

m/m = ( 50.0 / 50.0 + 150.0 )

m/m = ( 50.0 / 200 )

m/m = 0.25
8 0
3 years ago
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Help!!! I need the answer for "explain how using the periodic table can help answer this question"
Tanzania [10]
Well a compound is 2 or more different molecules together. So if you see a periodic table you will see the elements to see if they are the same or not.
3 0
2 years ago
If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0
3 years ago
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