4. What are the possible genotypes of the pes plant?

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

4 0

2 years ago

Difference between practical work inside and outside laboratory

arsen [322]

Practical work refers to the art of conducting experiments in order to answer certain research questions.

<h3>What is practical work?</h3>

In science, practical work refers to the art of conducting experiments in order to answer certain research questions. This could occur in a laboratory under controlled conditions or in the field.

In the physical sciences, most of the practical work is conducted in the laboratory under controlled conditions. However, some experiments in the biological sciences and most experiments in the social sciences are conducted outside the laboratory.

Learn more about experiments:brainly.com/question/11256472

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5 0

1 year ago

Because they have aromas, compounds with a ring of resonance bonds are called –

Shalnov [3]

Answer:

the answer is Fragrent compounds

6 0

3 years ago

Hydrophobic compounds dissolve in

Anastaziya [24]

Hydrophobic molecules tend to be nonpolar molecules that group together to form micelles rather than be exposed to water. Hydrophobic molecules typically dissolve in nonpolar solvents (e.g., organic solvents).

4 0

2 years ago

What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0

3 years ago