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3 years ago

What may happen to the human body when exposed to an infectious agent?

Chemistry

1 answer:

Hatshy [7]3 years ago

3 0

Answer:

When the body is exposed to viruses, bacteria, fungi, or parasites through an infection or vaccination the immune system creates antibodies and immune cells that inactivate or destroy the specific infectious organism.

Explanation:

hope this helps :D

You might be interested in

1.80x1024 molecules of CO2 is equal to how many grams?

olga nikolaevna [1]

Answer:

131.5177 grams

Explanation:

You divide by Avogadro's number to turn it into moles then multiply by molar mass.

7 0

3 years ago

What is the mass of a silver coin that contains 1.69x10^23 silver atoms? Report your answer in units of grams

Margaret [11]

Answer: 30.27 g

Explanation:

Since the silver is in units of atoms, we have to use Avogadro's number and molar mass.

Avogadro's number: 6.022×10²³ atoms/mol

Molar Mass: 107.87 g/mol

Now that we have everything we need, we can convert atoms to grams.

$1.69*10^{23} atoms*\frac{1mol}{6.022*10^2^3 atoms} *\frac{107.87 g}{1mol} =30.27 g$

3 0

3 years ago

Name the producer in the food chain? <br>

yKpoI14uk [10]

Answer:

Producers are autotrophs, or organisms that produce their own food. ... They are at the bottom of the food chain because they are eaten by other organisms, and they don't need to eat for energy. Producers make their own food through the process of photosynthesis instead of eating organic matter.

6 0

3 years ago

How many grams of carbon are contained in one mole of<img src="https://tex.z-dn.net/?f=C_%7B3%7DH_%7B8%7D" id="TexFormula1" titl

NNADVOKAT [17]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \:36 \:\: g$

____________________________________

$\large \tt Explanation \: :$

The given compound has 3 carbon atoms, so in 1 mole of that compound, there will be 3 moles of carbon atoms.

Mass of each mole of carbon atoms = 12 g

For 3 mole carbon atoms, it will be 12 × 3 = 36 grams

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0

2 years ago

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago