When you test starch with Barfoed’s reagent, what would be the answer, positive or negative?

A corpse discovered in the desert at 5:00 PM was found to have larvae from the blow fly species Phormia regina that had just dev

horrorfan [7]

Answer:

12.19 hours prior to discovery was the corpse placed in the desert.

Explanation:

Heat required for fly eggs to develop into first instar larvae = $Exposure\,time\times temperature$

From the given,

Exposure time = 16 hrs

Temperature = 27 C

$=16\times 27 = 427.2^{o}C$

The day time exposed hours is 10 hrs.-(7.00am-5.00pm)

$=10\times 37.8 = 378^{o}C$

So, the extra heat = 427.2-378= 49.2

The addition heat divided by average temperature.

Average temperature = 22.4 C

$=\frac{49.2}{22.4}= 2.19 hrs$

So, the total time exposed is night time hours to day time hours.

= 10+2.19 = 12.19 hrs.

Therefore, 12.19 hours prior to discovery was the corpse placed in the desert.

3 0

3 years ago

Which cell structure is correctly paired with its primary function?

Butoxors [25]

Answer:

C

Explanation:

Mitochondria provides energy (ATP) for cells to use

Nucleus provides genetic code (DNA)

Ribosomes assemble amino acids chains based on the DNA from nucleus to form proteins

6 0

3 years ago

Which of the following statements is true?

zalisa [80]

It's A, tThe figure is a molecule and an element.

6 0

3 years ago

Read 2 more answers

What Type of reaction is this equation C2H5S + O2 ---> CO2 + H2O + SO2

sweet-ann [11.9K]

I think it might be a decomposition.

4 0

3 years ago

Read 2 more answers

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

The activation energy for this reaction = 23 kJ/mol.

6 0

3 years ago