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3 years ago

Polymer formed from methylene diphenyl isocyanate

Chemistry

1 answer:

marin [14]3 years ago

5 0

Answer:

Polymeric MDI is a mixture of. monomeric MDI as well as larger molecular weight oligomers of MDI, and is a brownish. liquid at room temperature and may have a slight odor. Commercial MDI products are. often mixtures of monomeric and polymeric MDI and can contain other additives as well.

Explanation:

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Combustion of hydrocarbons such as nonane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp

maksim [4K]

Answer:

Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: Volume of CO₂ produced = 1223.21 L

Note: the complete second part of the question is given below:

2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

Explanation:

Part 1: Balanced chemical equation

C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: volume of carbon dioxide produced

From the equation of the reaction;

At s.t.p., I mole of C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂

molar mass of C₉H₂₀ = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L

Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.

O.470 Kg of nonane = 470 g of nonane

470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂

Volume of CO₂ gas produced at 1 atm and 17 °C;

Using P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁

where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K

Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)

Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂

3 0

4 years ago

26. All atoms area. positively charged, with the number of protons exceeding the number of electronsb, negatively charged, with

user100 [1]

(26) All atoms area...with the number of protons equaling the number of electrons

(27) The particles that are found in the nucleus of an atom are... protons and neutrons.

(28) As a consequence of the discovery of the nucleus by Rutherford, which model of the atom is thought to be true?...Protons. electrons, and neutrons are evenly distributed throughout the volume of the atom.

(29) The nucleus of an atom is...the central core and is composed of protons and neutrons.

7 0

3 years ago

Sea floor spreading occurs at a mid-ocean ridge true or false

Elenna [48]

Answer:

False

Explanation:

sea floor spreading is not consistent at all mid ocean ridges.

7 0

3 years ago

What is the electronic structure of carbon

horsena [70]

Answer:

[He] 2s2 2p2

Explanation:

5 0

4 years ago

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the balanced equation 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

4 years ago