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3 years ago

Where does air enter the body? A. nose B. mouth C. nose and mouth D. neither nose nor mouth

Chemistry

2 answers:

bearhunter [10]3 years ago

8 0

Nose and mouth so it would be C

fredd [130]3 years ago

6 0

The answer is C. Nose and mouth

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Please help soon thanks

avanturin [10]

Hydrosphere & Geosphere hope this helps
A and C

8 0

2 years ago

What is the initial temperature (°C) of a system that has the pressure decreased by 10 times while the volume increased by 5 tim

Ede4ka [16]

Answer:

300K

Explanation:

Given pressure of the system decreased by 10 times which means $P_{2} =\frac{P_{1} }{10}$

Given the volume of the system increased by 5 times which means $V_{2} =5\times V_{1}$

Given final temperature $T_{2}=150K$

Let the initial temperature be $T_{1}$

We know that PV=nRT

As n and R are constant $\frac{PV}{T}=constant$

$\frac{P_{1}V_{1} }{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}$

$T_{1} =\frac{P_1V_1}{P_2V_2} T_{2}$

$T_{1} =2\times T_{2}$

T1=300K

8 0

3 years ago

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$

Finally, the Gibbs free energy for the reaction at 298.15K is:

$\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol$

Best regards.

3 0

4 years ago

Alec saw footage of a landslide on the news. He saw rocks falling from the side of a mountain, with some of them smashing into e

viktelen [127]

Answer:

Explanation:

3 0

3 years ago

Within temperatures allowable for growth, higher temperatures usually lead to Choose one: A. The effect depends on the species.

ivolga24 [154]

Answer:

Explanation:

Higher temperatures leads to higher growth rates this is because, as the temperature increases, the plants have more needs to respire leading to more cellular activities and release of more carbonioxide. The increase in atmospheric carbondioxide and cellular activities leads to increase in plant growth. However, it should be noted that the increase in temperature shouldn't go beyond a certain level as it can also cause adverse effects of crops.

6 0

3 years ago