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yKpoI14uk [10]
3 years ago
14

Put in order please!!!!!! PLEASE!!!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
GuDViN [60]3 years ago
8 0
A. Groundwater
B. transpiration
C. Condensation
D. Interception
E. Infiltration
F. Evaporation
kondor19780726 [428]3 years ago
8 0
Its already in alphabetical order
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What is the pH of a 5.09 x 10-5 M solution of NaOH?
Varvara68 [4.7K]

<u>Answer:</u> The pH of the solution is 9.71

<u>Explanation:</u>

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions.

We are given:

pOH of the solution = 7.2

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOh of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=5.09\times 10^{-5}M

Putting values in above equation, we get:

pOH=-\log(5.09\times 10^{-5})\\\\pOH=4.29

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-4.29=9.71

Hence, the pH of the solution is 9.71

7 0
3 years ago
A book sits motionless on a table. Which statement is true? Group of answer choices There are no forces acting on the book. The
Katena32 [7]

Answer:

The forces acting on the book are balanced by each other.

6 0
2 years ago
How can one kg of iron melt more ice than 1 kg lead at 100 °C
Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{iron} = Heat obtained from the iron by the ice

ΔQ_{iron} = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{lead} = Heat obtained from the iron by the ice

ΔQ_{lead} = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of  ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0
3 years ago
C9H8O4 Aroms in formula
ikadub [295]
Yes it’s adokngd the formula one
6 0
3 years ago
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All solutions<br>are mixtures; but<br>all mixtures are<br>not solutions.<br>why?​
aleksklad [387]

All solutions are mixtures of two or more substances, but unless the mixture has a homogeneous distribution of solutes in the solvent, then the mixture is not a solution. Therefore, all mixtures are not solutions.

5 0
2 years ago
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