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Roman55 [17]
3 years ago
6

Forming a hypothesis is accomplished through___ reasoning

Chemistry
1 answer:
mixer [17]3 years ago
5 0
Scientific would be the word to fill in the blank
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Polymer formed from methylene diphenyl isocyanate
marin [14]

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:

5 0
3 years ago
For a phase change, H0 = 2 kJ/mol and A S0 = 0.017 kJ/(K•mol). What are
34kurt

Answer:

ΔG = -6.5kJ/mol at 500K

Explanation:

We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:

ΔG = ΔH - TΔS

Computing the values in the problem:

ΔG = ?

ΔH = 2kJ/mol

T = 500K

And ΔS = 0.017kJ/(K•mol)

Replacing:

ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)

ΔG = 2kJ/mol - 8.5kJ/mol

<h3>ΔG = -6.5kJ/mol at 500K</h3>

8 0
2 years ago
If you made 6 moles of NO2 How many grams of N2 did you use N2+2O2&gt; 2NO2​
denis23 [38]

Answer: 1:2

Explanation:

Believe me its correct.

7 0
2 years ago
What is the probability of a<br> short plant in the cross<br> between a Tt and a tt plant?
kondaur [170]
75% would be your answer since the lower letter represents a shorter plant
7 0
2 years ago
Calculate the theoretical carbonaceous and nitrogenous oxygen demand for:
serg [7]

Answer:

The correct answer is 129 mg and 232 mg.

Explanation:

Theoretical carbonaceous oxygen demand:

The reaction will be,  

C₂H₆O₂ + 5/2 O₂ ⇒ 2CO₂ + 3H₂O

Thus, for one mole of C₂H₆O₂ (ethylene glycol), 2.5 moles of O₂ is needed.  

The molecular mass of ethylene glycol is 62 grams per mole.  

The given mass of ethylene glycol is 100 mg or 0.1 grams

The moles of ethylene glycol will be,  

Moles = Weight/Molecular mass

= 0.1/62 = 1.613 × 10⁻³ mol

For 1.613 × 10⁻³ mol, the moles of O₂ will be,  

= 2.5×1.613×10⁻³

= 4.0.×10⁻³ × 32mol

= 0.129 grams or 129 mg.  

The theoretical nitrogenous oxygen demand is:  

The reaction will be,  

2NH₃-N + 9/2O₂ ⇒  4HNO2 + H₂O

Thus, for 2 moles of NH₃-N, 4.5 moles of O₂ is needed,  

Therefore, for 1 mol of NH₃-N, the oxygen required will be,  

= 4.5/2 = 2.25 mol

The given mass of NH₃-N is 100 mg, the moles of NH₃-N will be,  

Moles = 100×10⁻³/31 = 3.225 × 10⁻³ mol (The molecular mass of NH₃-N is 31 gram per mole)

The moles of O₂ is 2.25 × 3.225 × 10⁻³ = 7.258 × 10⁻³ mol.  

Now the mass of O2 will be,  

= 7.258 × 10⁻³ × 32

= 0.232 grams

= 232 mg

5 0
2 years ago
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