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Natasha2012 [34]
3 years ago
13

Convert 0.320 atm to mmhg

Chemistry
2 answers:
Molodets [167]3 years ago
5 0
<h2><em><u>PLEASE MARK BRAINLIEST</u></em></h2>

760 mmHg = 1 atm Therefore 30 atm = 760 * 3 mmHg = 2280 mmHg p1 = 700 mmHg p2 = 2280 mmHg You can compare them from there. You may also convert mmHg to atm by dividing by 760.

Your answer would be 243.2 mm Hg

Dmitry [639]3 years ago
3 0

Answer: 243.2 mmHg

Explanation:

Pressure is defined as the force per unit area. It is measured in units called as atmosphere , pascal , torr and mmHg.

The conversion factor is :

1atm=760mmHg

If 1 atmosphere corresponds to 760 mmHg

0.320 atm will correspond to = \frac{760}{1}\times 0.320=243.2mmHg.

Thus the pressure will be 243.2 mmHg

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6 0
3 years ago
Good morning, I have a question..
matrenka [14]

Answer:

<h3>The answer is 5.24 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 152 g

density = 29 g/cm³

We have

volume =  \frac{152}{29}  \\  = 5.24137931...

We have the final answer as

<h3>5.24 mL</h3>

Hope this helps you

7 0
3 years ago
What is the molarity when water is added to 4 moles of sodium chloride to make 0.5 liter of solution?
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       If    0.5 L of solution contains 4 mol
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                         x  = (4 mol · L)  ÷  (0.5 L)

                         x  = 8 mol

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Read 2 more answers
3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)
GrogVix [38]

Explanation:

(a)   The given reaction equation is as follows.

        Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g) (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}

Reduction-half-reaction: NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)

As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

      Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l) (acidic)

(b)   The given reaction equation is as follows.

        HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq) (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)

Oxidation-half reaction: Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

      4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}

Therefore, balancing the whole reaction equation in the basic medium as follows.

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6 0
3 years ago
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