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Verizon [17]
3 years ago
5

In 3–5 sentences, explain the various factors that should be considered when implementing green roofs

Chemistry
1 answer:
Kruka [31]3 years ago
6 0

Answer:

you should consider waterproofing. depending on the climate in your area, you need to apply multiple layers of waterproofing in the roof for it to support vegetation. you should also confused the types of plants. the types of plants is important for the success of your roof. if you live in a dry area, you don't want to choose plants that can live in dry areas without drying out.

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Which compound can be used to preserve biological specimens? A central C is double bonded to an O, and single bonded to 2 H. A c
Darina [25.2K]

Answer:

A central C is double bonded to an O, and single bonded to 2 H

Explanation:

Methanal, also known as formaldehyde, is a chemical compound used to preserve dead biological specimens for further study. It is called formalin when in a solution and it helps keep specimens in a fresh state by hardening the tissues of the specimen involved.

Formaldehyde is a gaseous compound that has an aldehyde functional group i.e. -CHO and has a chemical formula, H-CHO or CH2O as described in the question that a central C is double bonded to an O, and single bonded to 2 H (see attached image for structural formula).

4 0
4 years ago
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
icang [17]

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T_2 ),

P_1V_1 / T_1 = P_2V_2 / T_2,

T_2 = P_2V_2T_1 / P_1V_1,

T_2 = 0.73 atm * 1.8 L * 298.15 K / 1 atm * 1.2 L = ( 0.73 * 1.8 * 298.15 / 1 * 1.2 ) K = 326.47425 K,

T_2 = 326.47425 K = 53.32425 C

7 0
3 years ago
Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0L. What temperature must be maintained
Morgarella [4.7K]
Given:

P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?

Assume that the gas ideally, thus, we can use the ideal gas equation:

PV = nRT

where R = 0.0821 L atm/mol K

123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T

solve for T 

T = 295.72 K<span />
5 0
4 years ago
Read 2 more answers
Compare and contrast the concepts of average mass and relative mass​
Sphinxa [80]

Answer:

Relative and average atomic mass both describe properties of an element related to its different isotopes. However, relative atomic mass is a standardized number that's assumed to be correct under most circumstances, while average atomic mass is only true for a specific sample.

Explanation:

6 0
4 years ago
Can someone solve this problem 5
Westkost [7]

Answer:

2

Step-by-step explanation:

A. Moles before mixing

<em>Beaker I: </em>

Moles of H⁺ = 0.100 L × 0.03 mol/1 L

                   = 3 × 10⁻³ mol

<em>Beaker II: </em>

Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.

        H⁺][OH⁻] = 1 × 10⁻¹⁴   Divide each side by [H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/[H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)

             [OH⁻] = 0.01 mol·L⁻¹

Moles of OH⁻ = 0.100 L × 0.01 mol/1 L

                      = 1 × 10⁻³ mol

B. Moles after mixing

                 H⁺    +    OH⁻   ⟶ H₂O

I/mol:      3 × 10⁻³   1 × 10⁻³

C/mol:   -1 × 10⁻³  -1 × 10⁻³

E/mol:    2 × 10⁻³          0

You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.

You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.


C. pH

 [H⁺] = (2 × 10⁻³ mol)/(0.200 L)

        = 1 × 10⁻² mol·L⁻¹

 pH = -log[H⁺ ]

       = -log(1 × 10⁻²)

       = 2

6 0
3 years ago
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