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4 years ago

9

What career is when a person anlysis food in laboratory ?

1 answer:

Klio2033 [76]4 years ago

5 0

Hi!

Your Question,

what career is when a person anlysis food in laboratory ?

Answer,

Food makes the analysis that I know is trying to engineer food. That's called that in Turkey

Good Luck

#Turkey

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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

3 years ago

A solution is 40 cetic acid by mass. the density of this solution is 1.049 g/ml. Calculate the mass of pure acetic acid in 220 m

taurus [48]

The mass of pure acetic acid in 220 ml of the given solution at 20°C is 92.311 g

<h3>What is Acetic acid?</h3>

Acetic acid is a type of carboxylic acid and also known as ethanoic acid

Its formula is CH₃COOH.

It is an organic compound and is a colorless liquid

It is mostly used in the production of vinegar

40 % acetic acid by mass means,

40 g of acetic acid is dissolved in 100 g of solution.

The density of solution at 20°C,

$\rho = 1.049 g/ml$

We know,

$\rho = \frac{m}{V}$

$V = \frac{m}{\rho}$

The volume of the solution, $V = \frac{100}{1.049}$ = 95.33 ml

95.33 ml of solution contains 40 g of pure Acetic acid

220 ml of solution contains $\frac{40 \times 220}{95.33}$ = 92.311 g of pure Acetic acid

Thus, the mass of pure Acetic acid in 220 ml of solution at 20°C is 92.311 g

Learn more about acetic acid:

brainly.com/question/24304533

#SPJ4

7 0

2 years ago

Which of the following leads to a higher rate of diffusion?

miv72 [106K]

<h3><u>Answer;</u></h3>

Higher velocity of particles

<h3><u>Explanation;</u></h3>

The diffusion rate is determined by a variety of factors which includes;

Temperature such that the higher the temperature, the more kinetic energy the particles will have, so they will move and mix more quickly and the diffusion rate will be high.
Concentration gradient such that the greater the difference in concentration, the quicker the rate of diffusion.
Higher velocity of particles increases the diffusion rate as this means more kinetic energy by the particles and hence the particles will mix and move faster, thus higher diffusion rate.

5 0

3 years ago

Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The <u>equilibrium constant</u> is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

6 0

4 years ago

Compare molecular & covalent network crystal solid?

Daniel [21]

Answer:

Molecular solids and covalent network solids are two types of solid compounds. The key difference between molecular solid and covalent network solid is that <em>molecular solid forms due to the action of Van der Waal forces </em>where as <em>covalent network solid forms due to the action of covalent chemical bonds.</em>

hope this helps

8 0

2 years ago