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3 years ago

Características de la báscula granataria porfi

Chemistry

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

Lowkey dont know the language

Explanation:

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The decay of a living things allows chemical elements to be lost.

Lelu [443]

Answer: is is true

Explanation:

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1 year ago

Cytotoxic t cells can attack target cells with which chemical weapons?

Aneli [31]

Answer:

secrete cytotoxic substance which triggers apoptosis of target cell.

Explanation:

Cytotoxic T cells have cell surface receptor which recognizes the antigen present on the receptor of target cell. This interaction initiates the process of killing of target cell.

After interaction cytotoxic t cell release cytotoxic substance called granzyme and perforin. Granzyme triggers apoptosis through the activation of caspases or by making the release of cytochrome c and activation of the apoptosome.

Perforin make pores in the cell and its action is similar to complement membrane attack complex. Therefore cytotoxic substances are released by Tc cells which trigger apoptosis of target cell.

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2 years ago

Nitrogen has the atomic number 7. an isotope of nitrogen containing seven neutrons would be

BigorU [14]

An isotope of nitrogen containing 7 neutrons would be nitrogen-7

8 0

3 years ago

A urine sample has a mass of 147 g and a density of 1.020 g/mL. What is the volume of this sample?

son4ous [18]

So we know that the equation for density is:

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

where D is the density, m is the mass in grams, and V is the volume in mL.

So since we know two of the variables, mass and density, we can solve for the volume:

$\frac{1.020g}{mL}=\frac{147g}{V}$

$V=\frac{147g}{\frac{1.020g}{mL}}$

$V=144.12mL$

Therefore, the volume of this urine sample is 144.12mL.

5 0

3 years ago

A sample of paper from an ancient scroll was found to contain 39.5% 14C content as compared to a present-day sample. The t1/2 fo

nirvana33 [79]

Answer

7665 years

Procedure

Let N₀ be the amount of carbon-14 present in a living organism. According to the radioactive decay law, the number of carbon-14 atoms, N, left in a dead tissue sample after a certain time, t, is given by the exponential equation:

N = N₀e^(-λt)

where λ is the decay constant which is related to half-life (T1/2) by the equation:

$\lambda=\frac{ln(2)}{t_{\frac{1}{2}}}$

Here, ln(2) is the natural logarithm of 2.

The percent of carbon-14 remaining after time t is given by N/N₀.

Using the first equation, we can determine λt.

The half-life of carbon-14 is 5,720 years, thus, we can calculate λ using the second equation, and then find t.

$\lambda=\frac{ln(2)}{5720}=1.211\times10^{-4}$

Solving the second equation for t, and using the λ we have just calculated we will have

t= 7665 years

3 0

9 months ago