43 inHg = 43 inHg*2.54cm/in = 109.22cmHg * 10 mm/cm = 1092.2 mmHg

14.7 psi = 760 mmHg

1092.2mmHg * 14.7psi / 760 mmHg = 21.13 psi

Answer: option D. 21.13 psi

7 0

3 years ago

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Which is not a descriptor or tactic used by abusers?

attashe74 [19]

It should be B. All of the other choices are used often by abusers

4 0

3 years ago

nalin [4]

Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.

6 0

2 years ago

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At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank

hodyreva [135]

Answer:

$\large \boxed{\text{-92 $^{\circ}$C}}$

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

$\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}$

2. Temperature in degrees Celsius

$\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}$

8 0

2 years ago