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2 years ago

11

rain, freezing rain, sleet, snow, and hail are all examples of a. infiltration. b. precipitation. c. evaporation. d. condensatio

n.

2 answers:

Fudgin [204]2 years ago

6 0

The correct answer is precipitation.

Vladimir79 [104]2 years ago

5 0

These are all examples of B. Precipitation.

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Some atoms are chemically stable and will not bond with atoms of other

tigry1 [53]

Answer:

The stability of atoms depends on whether or not their outer-most shell is filled with electrons. If the outer shell is filled, the atom is stable. Atoms with unfilled outer shells are unstable, and will usually form chemical bonds with other atoms to achieve stability.

Explanation:

7 0

2 years ago

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PLEASE HELP!?!.!,! What are the products of the combustion of a hydrocarbon?

Tatiana [17]

Answer:

carbon dioxide and water

Explanation:

Example: Combustion of Methane (CH₄(g))

CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**

____________________

Note: The combustion of any hydrocarbon produces CO₂ & H₂O. That is,

Ethane (C₂H₆) + O₂ => CO₂(g) + H₂O(g)

Propane (C₃H₈) + O₂ => CO₂(g) + H₂O(g)

Butane (C₄H₁₀) + O₂ => CO₂(g) + H₂O(g)

The issue remaining is to balance the reaction equation. For these type equation balance Carbon 1st, then Hydrogen and finish with Oxygen. Balancing in this order leaves Oxygen which can be balanced using fractions. If problem requires lowest whole number ratios of elements, simply multiply entire equation by 2 to get standard equation*

______________________

*Standard Equation is defined as the smallest whole number ratios of elements. The 'standard equation' is significant in that it is assumed to be at STP conditions; i.e., 0⁰C (=273K) & 1.0 Atmosphere pressure.

Ethane (C₂H₆) + 7/2O₂(g) => 2CO₂(g) + 3H₂O(g)

=> 2C₂H₆ + 7O₂(g) => 4CO₂(g) + 6H₂O(g) <= Standard Form of Rxn

Propane (C₃H₈) + 5O₂(g) => 3CO₂(g) + 4H₂O(g) <= Standard Form of Rxn (no need to balance with the '2' multiple)

Butane (C₄H₁₀) + 13/2O₂ => 4CO₂(g) + 5H₂O(g)

=> 2C₃H₈ + 13O₂(g) => 4CO₂(g) + 5H₂O(g) <= Standard Form of Rxn

______________________

**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).

3 0

3 years ago

Read 2 more answers

GarryVolchara [31]

Answer:

The value of the Golden Igloo is $227.4 million.

Explanation:

First, we need to find the inner and the outer volume of the half-spherical shell:

$V_{i} = \frac{1}{2}*\frac{4}{3}\pi r_{i}^{3}$

$V_{o} = \frac{1}{2}*\frac{4}{3}\pi r_{o}^{3}$

The total volume is given by:

$V_{T} = V_{o} - V_{i}$

Where:

$V_{i}$ : is the inner volume

$r_{i}$ : is the inner radius = 1.25/2 = 0.625 m

$V_{o}$ : is the outer volume

$r_{o}$ : is the outer radius = 1.45/2 = 0.725 m

Then, the total volume of the Igloo is:

$V_{T} = \frac{2}{3}\pi r_{o}^{3} - \frac{2}{3}\pi r_{i}^{3} = \frac{2}{3}\pi [(0.725 m)^{3} - (0.625 m)^{3}] = 0.29 m^{3}$

Now, by using the density we can find the mass of the Igloo:

$m = 19.3 \frac{g}{cm^{3}}*0.29 m^{3}*\frac{(100 cm)^{3}}{1 m^{3}} = 5.60 \cdot 10^{6} g$

Finally, the value (V) of the antiquity is:

$V = \frac{\$ 1263}{oz}*5.60 \cdot 10^{6} g*\frac{1 oz}{31.1034768 g} = \$ 227.4 \cdot 10^{6}$

Therefore, the value of the Golden Igloo is $227.4 million.

I hope it helps you!

8 0

2 years ago

Calculate the speed of sound wave with a frequency of 20hz and a wave length of 3.00m

Marta_Voda [28]

Wavelength is the distance of one frequency wave peak to the other and is most commonly associated with the electromagnetic spectrum.[1] Calculating wavelength is dependent upon the information you are given. If you know the speed and frequency of the wave, you can use the basic formula for wavelength. If you want to determine the wavelength of light given the specific energy of a photon, you would use the energy equation. Calculating wavelength is easy as long as you know the correct equation.

7 0

2 years ago

Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho

AleksandrR [38]

The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
$(1/2) ^{n} =x$
$t_{1/2} = \frac{t}{n}$
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span> $t_{1/2}$ - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span> $(1/2) ^{n} =x$
</span> $(0.5) ^{n} =0.125$
$n*log(0.5)=log(0.125)$
$n= \frac{log(0.5)}{log(0.125)}$
$n=3$

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span> $t_{1/2} = 12.4$
</span><span> $t_{1/2} *n = t$
</span> $t= 12.4*3$

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>

7 0

3 years ago

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