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mote1985 [20]
3 years ago
14

A sample of nitrogen monoxide occupies a volume of 400 mL at 100 °C and 720 mmHg.

Chemistry
1 answer:
konstantin123 [22]3 years ago
6 0
The volume of nitrogen  monoxide that   occupy  at  STP  is=  277  Ml

     calculation
 The volume is obtained  using   the combined  gas law  that is P1V1/T1= P2V2?
Where  P1 = 720  MmHg
            V1 = 400ml
           T1= 100 +273=  373 K
At  STP  temperature  =  273 K  and  pressure=  760 mm  hg
therefore T2=  273 k
               p2 = 760  mmhg
               V2=?
make  V2  the subject of the formula

V2= (T2 ×P1 ×V1)/(P2×T1)

V2  is therefore  = (273k  x720 mmhg x 400 ml)/(760 mmhg x373K) = 277  Ml
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Remembering that
d = m ÷ v

d = ?
m = 104 g
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<span>d = 4,95 g÷cm³</span>
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A small piece of Cr metal reacts with dilute HNO3 to form H2 (g), which is collected over water at 18 C in a large flask. The to
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The partial pressure is P_p= 737.5 mm \ of  \ Hg

Explanation:

The Partial pressure of H_2 is mathematically represented as

           P_p = P_T -P_w

Where P_T is the total pressure of water with a value of 15.5 mm of Hg

           P_w is the partial pressure of water with a value 753 mm of Hg

Now substituting values

          P_p = 753-15.5

               P_p= 737.5 mm \ of  \ Hg

       

3 0
3 years ago
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What is happening at the atomic level to give rise to the observed energies in the flame tests and in the emission spectra
irina1246 [14]

Answer:

See explanation

Explanation:

According to the Bohr model of the atom, electrons are found in energy levels. Energy is absorbed or emitted when an electron moves from one energy level to another.

During flame test, electrons absorb energy and move to higher energy levels; they quickly return to ground state and emit the energy previously absorbed as a photon of light. This is seen as the colour imparted to the flame by the metal.

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If wind is eroding an area at a rate of 2 mm per year and depositing it in a smaller area at a rate of 7 mm per year, how much l
amid [387]

Answer:

After 2000 years, the first after will be 4m lower.

After 2000 years, the second area will be 14m higher.

Explanation:

From the given information:

At the first area:

if the wind is eroding at 2 mm/year, then after 2000 years; we have:

2 mm/year × 2000 year = 4000 mm

4000 mm to meters, we have:

=4000 \ mm \times \dfrac{1 \ m}{ 1000 \ mm}

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At the second smaller area:

Given that the wind is depositing it at the rate of 7 mm per year;

Then:

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6 0
3 years ago
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