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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

brainly.com/question/19694949

#SPJ1

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Crazy boy [7]
<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>

Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.

Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3

To find for the theoretical yield, we first determine the limiting reactant.

100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2

Therefore, the limiting reactant is O2.

Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3

Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
3 0
3 years ago
A student weighs out 3.68 grams of cobalt (II) chloride to do an experiment with. How many moles of is this?
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Answer:

0.029mol

Explanation:

from

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(Hurry ASAP!!!) There are 1.5 moles of methane (CH4) in a tank. It has a pressure of 4.4 atm at 173°C. Find the volume occupied
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Answer:

hope this help !

Explanation:

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What is the molarity of aqueous lithium bromide if 25.0 mL of LiBr reacts with 10.0 mL of 0.250 M Pb(NO 3) 2
bixtya [17]

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We'll begin by calculating the number of mole of Pb(NO₃)₂ in the solution.

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  • Molarity of Pb(NO₃)₂ = 0.250 M
  • Mole of Pb(NO₃)₂ =?

Mole = Molarity x Volume

Mole of Pb(NO₃)₂ = 0.25 × 0.01

Mole of Pb(NO₃)₂ = 0.0025 mole

Next, we shall determine the mole of LiBr required to react with 0.0025 mole of Pb(NO₃)₂

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From the balanced equation above,

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Molarity = mole / Volume

Molarity of LiBr = 0.005 / 0.025

Molarity of LiBr = 0.2 M

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