Which objects will most likely float

Chemistry

1 answer:

storchak [24]4 years ago

6 0

Sponges, apples, light weight balls, etc.
they’re less dense than water

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The partial pressure of cartbon dioxide in the atmgsphere is 0239 toer Caiculate the partial pressure in mm He and atm Rpund eac

Verizon [17]

Answer :

The pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

The pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

Explanation :

The conversion used for pressure from torr to mmHg is:

1 torr = 1 mmHg

The conversion used pressure from torr to atm is:

1 atm = 760 torr

or,

$1torr=\frac{1}{760}atm$

As we are given the pressure of carbon dioxide in the atmosphere 0.239 torr. Now we have to determine the pressure of carbon dioxide in the atmosphere in mmHg and atm.

Pressure in mmHg :

As, $1torr=1mmHg$

So, $0.239torr=\frac{0.239torr}{1torr}\times 1mmHg=0.239mmHg$

Thus, the pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

Pressure in atm:

As, $1torr=\frac{1}{760}atm$

So, $0.239torr=\frac{0.239torr}{1torr}\times \frac{1}{760}atm=3.14\times 10^{-4}atm$

Thus, the pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

3 0

3 years ago

A pump contains 1.5 L of air at 175 kPa. You draw back on the piston of the pump, expanding the volume until the pressure reads

Tresset [83]

Answer:

$\large \boxed{\text{86.8 L}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rclrcl}p_{1}& =& \text{0.579 atm}\qquad & V_{1} &= & \text{150 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}0.579 \times 150 & =& 1.00V_{2}\\86.85 & = & 1.00V_{2}\\V_{2} & = &\dfrac{86.85}{1.00}\\\\& = &\textbf{86.8}\\\end{array}\\\text{The new volume of the gas is } \large \boxed{\textbf{86.8 L}}$