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olganol [36]
3 years ago
13

the pressure of a gas in a container is 152 mm hg. This is equivalent to _____. 0.300 atm 2.00 atm 0.400 atm 0.200 atm

Mathematics
2 answers:
kolbaska11 [484]3 years ago
8 0

The pressure of a gas in the container is 152 mm hg.

Now to convert this pressure of gas from 'mm hg' to 'atm', we need to follow the given conversion:

As we know that,

1 millimeter of mercury = 0.00131 atmospheric pressure

Since we know the value of 1 mm of hg = 0.00131 atm. So,  152 millimeter of mercury should be equal to:

152 \times 0.00131=0.19912

So, the pressure of the gas in the container is 0.19912 atmospheric pressure or 0.200 atm.

uranmaximum [27]3 years ago
5 0

Answer: 0.200 atm

Step-by-step explanation:

Since we know that One atm = 760 millimeters Hg

Then, \text{1 mm Hg}=\dfrac{1}{760}\text{ atm}

Now, \text{152 mm Hg}=\dfrac{152}{760}\text{ atm}

\text{1 mm Hg}=0.2\text{ atm}

Hence, the complete statement will be : The pressure of a gas in a container is 152 mm hg. This is equivalent to <u>0.200</u> atm .

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As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

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{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

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