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3 years ago

Correctly label the parts of the water cycle enter our answer in the space provided and include the question numbers

Chemistry

1 answer:

strojnjashka [21]3 years ago

3 0

The correct matches are:

1 - Evaporation;

This process is caused by the heat of the Sun. Basically, the water turns into gaseous form and because it is very light it goes up into the atmosphere.

2 - Condensation;

Once the water vapor reaches certain heights in the atmosphere, it starts to condensate. This means that the water vapor turns into liquid form again, becoming small drops of water and forming a cloud.

3 - Precipitation;

Once the drops of water are formed, under the influence of the gravity, they start to fall on the surface of the Earth. This is known as precipitation.

4 - Runoff;

Once the water from the precipitation reaches the land, the water moves from the higher toward the lower areas. This is know as runoff.

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3 years ago

Which of the following statements about the combustion of glucose with oxygen to form water and carbon dioxide (C6H12O6 + 6 O2 →

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Answer:

The correct statement is:

E - The entropy of the products is greater than the entropy of the reactants.

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3 years ago

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. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

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3 years ago

For the following reaction, 6.94 grams of water are mixed with excess sulfur dioxide . Assume that the percent yield of sulfurou

Alexxx [7]

<h3>Answer:</h3>

#a. Theoretical yield = 31.6 g

#b. Actual yield = 25.72 g

<h3>Explanation:</h3>

The equation for the reaction between sulfur dioxide and water to form sulfurous acid is given by the equation;

SO₂(g) + H₂O(l) → H₂SO₃(aq)

The percent yield of H₂SO₃ is 81.4%

Mass of water that reacted is 6.94 g

#a. To get the theoretical yield of H₂SO₃ we need to follow the following steps

Step 1: Calculate the moles of water

Molar mass of water = 18.02 g/mol

Mass of water = 6.94 g

But, moles = Mass/molar mass

Moles of water = 6.94 g ÷ 18.02 g/mol

= 0.385 mol

Step 2: Calculate moles of H₂SO₃

From the equation, the mole ratio of water to H₂SO₃ is 1 : 1

Therefore, moles of water = moles of H₂SO₃

Hence, moles of H₂SO₃ = 0.385 mol

Step 3: Theoretical mass of H₂SO₃

Mass = moles × Molar mass

Molar mass of H₂SO₃ = 82.08 g/mol

Number of moles of H₂SO₃ = 0.385 mol

Therefore;

Theoretical mass of H₂SO₃ = 0.385 mol × 82.08 g/mol

= 31.60 g

Thus, the theoretical yield of H₂SO₃ is 31.6 g

<h3>#b. Calculating the actual yield</h3>

We need to calculate the actual yield

Percent yield of H₂SO₃ is 81.4%

Theoretical yield is 31.60 g

But; Percent yield = (Actual yield/theoretical yield)×100

Therefore;

Actual yield = Percent yield × theoretical yield)÷ 100

= (81.4 % × 31.6) ÷ 100

= 25.72 g

The percent yield of H₂SO₃ is 25.72 g

6 0

3 years ago