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Oksana_A [137]
2 years ago
13

the roof of a building is 0.2km^2. during a rainstorm, 5.5 cm of rain was measured to be sitting on the roof. what is the mass i

n kg of the water on the roof after the rainstorm? (density of rainwater = 1g/ml)
Chemistry
1 answer:
Nikolay [14]2 years ago
7 0

The mass in kg of the water on the roof after the rainstorm is mathematically given as

The mass of the water on the roof after a rainstorm is 1.1 *10^{10}g  or 1.1*10^{7}kg

<h3>What is the mass in kg of the water on the roof after the rainstorm? </h3>

Generally, the equation for the Area of the roof is  mathematically given as

A=0.20 \times 10^{10} \mathrm{~cm}^{2}

height of the rain =5.5cm

the volume of the rain on the roof =1.1 * 10^{10} CC

Generally, the equation for mass is  mathematically given as

mass=volume*density

&=\left(1.1 \times 10^{10} \mathrm{cC} \times 1 \mathrm{~g} / \mathrm{cc}\right) \\&=1.1 \times 10^{10} \mathrm{~g}

In conclusion, the Mass of the water on the roof after a rainstorm is 1.1 *10^{10}  or $1.1*10^{7}

Read more about Mass

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olga2289 [7]

Answer: The empirical formula for the given compound is NO_2

Explanation : Given,

Mass of O = 0.370 g

Mass of N = 0.130 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.370g}{16g/mole}=0.0231moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.130g}{14g/mole}=0.00928moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00928 moles.

For Oxygen  = \frac{0.0231}{0.00928}=2.4\approx 2

For Nitrogen = \frac{0.00928}{0.00928}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of O : N = 2 : 1

Hence, the empirical formula for the given compound is NO_2

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3 years ago
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mash [69]

Answer:

The classification is mentioned below for the particular topic.

Explanation:

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⇒      3Zn+2H_{3}Po_{4}\rightarrow Zn_{3}(Po_{4})_{2}+3H_{2}

  • Since fresh zinc complicates the cycle since, as a comparison to polluted zinc, there was little contact with either the reaction.
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consider an atom having four distinct neergy levels. if an electron is able to make transitions between, any two levels, how man
ludmilkaskok [199]

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From energy level 1 to 2 is one frequency, from energy level 1 to 3 is one frequency and From energy level 1 to 4 is one frequency. So, we have a total of 3 frequencies for transition from energy level 1.

From energy level 2 to 3 is one frequency and from energy level 2 to 4 is one frequency. So, we have a total of 2 frequencies for transition from energy level 2.

From energy level 3 to 4 is one frequency.

So we have a total of 3 + 2 + 1 different frequencies = 6 different frequencies.

Note that the reverse process for each step produces the same frequency as the step in consideration.

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A sample of the chiral molecule limonene is 62% enantiopure. what percentage of each enantiomer is present? what is the percent
ikadub [295]

The mixture contains 62 % one isomer and 38 % the enantiomer.

Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.

Then % (<em>S</em>) = 100 % -62 % = 38 %

ee = %  (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %


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3 years ago
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I: Current
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