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muminat
3 years ago
11

True or false? Even when soil is saturated with water, water is still available in small spaces.

Chemistry
1 answer:
kirill [66]3 years ago
8 0
This is True.........
You might be interested in
2.56 g of hydrogen reacts completely with 20.32 g of oxygen<br> to form X g of water. X = g
Brilliant_brown [7]

Answer:

Mass of water produced is 22.86 g.

Explanation:

Given data:

Mass of hydrogen = 2.56 g

Mass of oxygen = 20.32 g

Mass of water = ?

Solution:

Chemical equation:

2H₂ + O₂   →  2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 20.32 g/ 32 g/mol

Number of moles = 0.635 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 2.56 g/ 2 g/mol

Number of moles = 1.28 mol

Now we will compare the moles of water with oxygen and hydrogen.

                    O₂            :            H₂O

                     1              :             2

                  0.635        ;            2×0.635 =  1.27

                   H₂             :              H₂O

                    2              :              2

                 1.28            :           1.28

The number of  moles of water produced by oxygen are less thus it will be limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 1.27 × 18 g/mol

Mass = 22.86 g

 

4 0
3 years ago
Read 2 more answers
Which statement best describes a typical difference that could be found between the "Analysis" and "Conclusion" sections of a
IgorC [24]

The statement “Only the “Conclusion” section discusses whether the original hypothesis was supported, and both sections suggest further  research”, best describes the difference between analysis and conclusion.

Answer: Option 4

<u>Explanation: </u>

In research, we do experiments and derive the results. Then, those results were analyzed by us. In this analysis part, we compare our results with the related results published elsewhere. Also, we correlate the similarities and point out the differences between our analysis and other reported results.

In conclusion part, we have to check hypothesis or it supported. And, we summarise our analysis and figure out the further research need to be done on that to improvise our research. So, the final statement is the correct option which best describes the difference between analysis and conclusion.

3 0
3 years ago
An unknown compound contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 2.77 g sample of the compound
alisha [4.7K]

The molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

<h3>Molecular formula of the compound</h3>

The molecular formula is calculated as follows;

CHNO  +  O₂ ------------> CO₂  + H₂O

Mass of carbon, C:  = (6.64 x 12)/44 = 1.81 g in 2.77 g sample

Mass of hydrogen, H: = (1.67 x 2)/18 = 0.186 g in 2.77 g sample

Mass of Nitrogen, N: = (2.77 x 0.143)/1.69 = 0.234 g

Mass of oxygen, O:  = 2.77 g - 1.81 g - 0.186 g - 0.234 g = 0.54 g

<h3>molar ratio of the elements: </h3>

C = 1.81 g = 0.15 mol

H = 0.186 g = 0.186 mol

N = 0.234 g = 0.017 mol

O = 0.54 g = 0.0337 mol

divide through with the smallest number of moles (0.017 mol);

C = 9

H = 11

N = 1

O = 2

Molecular formula = C₉H₁₁NO₂

Check the molar mass of the compound = (9 x 12) + (11 x 1) + (14) + (2 x 16) = 165 g/mol

Thus, the molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

Learn more about molecular mass here: brainly.com/question/21334167

#SPJ1

4 0
2 years ago
The iupac name of this ??
Licemer1 [7]
Methylhexanamine<span> and its formula is C7H17N</span>
that thing is a drug
3 0
3 years ago
A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i
In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>

<em />

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

<em>Moles CaCl₂ = Moles CaCO₃:</em>

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0
3 years ago
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