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3 years ago

6-11don't answer if you don't know

Chemistry

1 answer:

Bumek [7]3 years ago

6 0

Answer:

ghrtdfghtyfrhjbfyrt

Explanation:

t6n yrtnyrt tryrbhyrty ryrhtr

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Chemical barriers include? mucous food hydrochloric acid sweat

mr Goodwill [35]

Enzymes that eventually break down the surface of bacteria are found in saliva, mucus, and tears, which act as chemical barriers against infection.

Enzymes found in saliva, mucous, as well as tears that degrade bacterial surfaces serve as chemical barriers preventing infection. The stomach's acid as well as the acid in sweat both destroy cellular pathogens, and even the semen contains antibacterial proteins.

There is no chemical and physical barrier offered by macrophages. Instead of acting as a barrier, macrophages target pathogens once they have already entered the body. Physical barriers include mucus as well as stratified squamous epithelium, whereas chemical barriers include saliva and urine.

Therefore, Enzymes that eventually break down the surface of bacteria are found in saliva, mucus, and tears, which act as chemical barriers against infection.

To know more about chemical barriers

brainly.com/question/20725429

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4 0

1 year ago

The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C

Kipish [7]

Answer: 72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Putting in the values we get:

$\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles$

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

According to stoichiometry:

1 mole of $C_3H_8$ requires 5 moles of oxygen

0.45 moles of $C_3H_8$ require= $\frac{5}{1}\times 0.45=2.25$ moles of oxygen

Mass of $O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g$

72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

7 0

3 years ago

Which one would it be <br> 1,2,3,4

sesenic [268]

I think it's 1. PCl5= PCl3 + Cl2. I can't tell the exact reason but it's something to do with Atomic Redox Reaction

6 0

3 years ago

In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the high-temp

Paul [167]

Solution: The given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

If we have a hypothetical equation:

$A+2B\rightarrow 3C+5D$

Then the rate could be written as:

$rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}$

From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.

Like the above example, we can write the rate for the given equation and it would be looking as:

$rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}$

Now we can easily answer all the parts of the question.

(a) From above expression, the rate of consumption of $O_2$ related to rate of consumption of $NH_3$ as:

$-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

multiply both sides by -5

$\frac{\Delta [O_2]}{\Delta t}=\frac{5}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, rate of consumption of oxygen is $\frac{5}{4}$ the rate of consumption of ammonia.

(b) The relationship between rate of formation of NO to the rate of consumption of ammonia will be written as:

$\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 4

$\frac{\Delta [NO]}{\Delta t}=-\frac{\Delta [NH_3]}{\Delta t}$

So, rate of formation of NO equals to the rate of consumption of ammonia.

Now, the rate of formation of $H_2O$ to the rate of consumption of ammonia would be:

$\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 6

$\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, the rate of formation of $H_2O$ is $\frac{6}{4}$ times that is 1.5 times to the rate of consumption of ammonia.

5 0

4 years ago

Explain how the fire spreading from tree to tree provides a model for conduction.

Inessa05 [86]

the fire spreading is what represents the form of heat transferring from one atom to another within an object and direct contact

6 0

3 years ago

6-11don't answer if you don't know​

6-11don't answer if you don't know