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cluponka [151]
2 years ago
11

What type of soil holds the least amount of water

Chemistry
2 answers:
jeka57 [31]2 years ago
8 0
The clay soil holds the least amount if water.
dem82 [27]2 years ago
6 0
Sand soil is has the least  water hope this helped


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KOH(Ac) + HNO3 → KNO3(Ac) + H2O<br> cual es su reacio
Bumek [7]

Answer:

corbonization for dictionnal

Explanation:

correct

6 0
2 years ago
Read 2 more answers
The actual number of pennies in a jar is 218. You miscounted and came up with 215 pennies. What is your percent error?
Katyanochek1 [597]

Answer:

3%

Explanation:

Substract the actual error from the final and multiply by 100

6 0
2 years ago
Help??
kvasek [131]

Answer:

Mass of sodium chloride decomposed = 24.54 g

Explanation:

Given data:

Mass of sodium chloride decomposed = ?

Mass of chlorine gas formed = 15 g

Solution:

Chemical equation:

2NaCl      →         2Na + Cl₂

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 15 g/ 71 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of Cl₂ with NaCl from balance chemical equation.

                    Cl₂            :              NaCl

                      1              :                2

                      0.21         :            2×0.21 = 0.42 mol

Mass of Sodium chloride decompose:

Mass = number of moles × molar mass

Mass = 0.42 mol × 58.44 g/mol

Mass = 24.54 g

4 0
2 years ago
Please help. balance this equation
Nuetrik [128]

Answer:

k=2

ci=2

o=6

k=2

ci=2

o=6

Explanation:

4 0
2 years ago
I need this answer quick please show work
Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

4 0
3 years ago
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