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atroni [7]
3 years ago
10

Water circulates through Earth's water cycle by changing state state its acidity its acidity its turbidity

Chemistry
1 answer:
Lana71 [14]3 years ago
3 0

Complete question is;

Water circulates through Earth's water cycle by changing its _____.

A. state

B. acidity

C. turbidity

D. composition

Answer:

A: State

Explanation:

We can talk about this by observations about the states of water in the Earth's water cycle and even nature.

For example, water changing from liquid to steam(gas) in evaporation; under certain temperatures/conditions snow undergoes sublimation directly from solid to gas and also undergoes melting from solid to liquid; water vapour in the atmosphere undergoes condensation and cooling to form rain.

Thus, we can say that water circulates through the Earth's water by changing it's state.

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Answer:

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When one substance dissolves into another, a solution is formed. A solution is a homogeneous mixture consisting of a solute dissolved into a solvent

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3 years ago
How does the law of conservation of mass apply to this reaction
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8 0
3 years ago
Which equation is set up correctly to determine the volume of a 3.2 mole sample of oxygen gas at 50°C and 101.325 kPa?
77julia77 [94]
N = 3.2 moles, T = 50 + 273 = 323 K, P = 101.325 kPa,  R = 8.314 L.kPa/K.mol 

PV = nRT

V = nRT / P           substituting.

V = (3.2 mole)(8.314 L.kPa/K.mol )(323 K) / (<span>101.325 kPa)

That is the answer, but it is not among the options you provided. Check your options properly.</span>
7 0
3 years ago
Read 2 more answers
To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water
Masteriza [31]

Answer:Sample Absorbance (625 nm)  

A 0.536  

B 0.783  

C 0.045  

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738  

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0
3 years ago
What is the value of delta Hrxn for this equation:
Ratling [72]

Answer:

The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

The reaction bond enthalpies are for the reactant;

3 × N-H = 3 × 390 = 1,170 kJ/mol

2 × O=O = 2 × 502 = 1004 kJ/mol

The reaction bond enthalpies are for the product;

3 × N-O = 3 × 201 = 603 kJ/mol

3 × O-H = 3 × 464 = 1,392 kJ/mol

The ΔHrxn for the above equation is therefore;

ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol

6 0
2 years ago
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