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Oksana_A [137]
1 year 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]1 year 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

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If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced
ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • P₄: 1 mole
  • H₂: 6 moles
  • PH₃:4 moles

Being the molar mass of the compounds:

  • P₄: 124 g/mole
  • H₂: 2 g/mole
  • PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

  • P₄: 1 mole* 124 g/mole= 124 g
  • H₂: 6 mole* 2 g/mole= 12 g
  • PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

  • P= 2 atm
  • V= 10 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 298 K

Replacing:

2 atm*10 L= n*0.082 \frac{atm*L}{mol*K} *298 K

and solving you get:

n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

  • If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3}  }{124grams of P_{4}}

moles of PH₃=0.2742

  • If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2}  }{124grams of P_{4}}

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

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