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What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.

ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_A+p_B+p_C...$

Given : $p_{total}$ =total pressure of gases = ?

$p_{O_2}$ = partial pressure of oxygen = 100 mm Hg

$p_{N_2}$ = partial pressure of nitrogen = 573 mm Hg

$p_{CO_2}$ = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

$p_{H_2O}$ = partial pressure of water vapor = 47 torr = 47 mm Hg (1torr=1 mm Hg)

putting in the values we get:

$p_{total}=(100+573+40.28+47)mmHg$

$p_{total}=760.28mmHg$

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

5 0

3 years ago

CH,O<br> Needed:<br> Available:<br> Shared:<br> Bonds:

Alona [7]

Answer:

hy do C-C bonds and C-H bonds have high potential energy levels? ... electronegativity so the electrons in the covalent bond are equally shared. Term. Which functional groups are commonly found on a carbohydrate? ... few organisms have the specific enzymes needed to break the bond in the B-form ...

Explanation:

3 0

2 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

Lead (II) nitrate reacts with ammonium carbonate to produce ammonium nitrate

vitfil [10]

Pb(NO₃)₂ + (NH₄)₂CO₃ → PbCO₃ + 2 NH₄NO₃

Explanation:

Reaction of lead (II) nitrate with ammonium carbonate will produce lead (II) carbonate and ammonium nitrate.

The balanced chemical equation is:

Pb(NO₃)₂ + (NH₄)₂CO₃ → PbCO₃ + 2 NH₄NO₃

To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.

Learn more about:

balancing chemical reactions

brainly.com/question/13911443

#learnwithBrainly

3 0

3 years ago

Why is nh3 classified as a polar molecule

Yanka [14]

Explanation:

An ammonia $(NH_{3})$ molecule contains electronegative nitrogen atom. The difference between electronegativity of nitrogen and hydrogen atom is high.

Therefore, nitrogen will attract the shared pair of electrons more towards itself. As a result, there will be partial positive charge on hydrogen and partial negative charge on nitrogen.

Thus, due to these charges $NH_{3}$ is polar in nature.

4 0

3 years ago