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S_A_V [24]
3 years ago
5

_P4 + ____O2 → __P2O3 How do I balance this equation?

Chemistry
2 answers:
enot [183]3 years ago
7 0

Explanation:

<em>hope</em><em> it's</em><em> helpful</em><em> for</em><em> you</em>

<em>pls</em><em> mark</em><em> above</em><em> guy</em><em> ans</em><em> as</em><em> brainliest</em>

Murljashka [212]3 years ago
5 0

Explanation:

please mark me as brainlest

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8 0
3 years ago
What element has greatest atomic radius from group 3?
stira [4]

Answer:

i think iron

Explanation:

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6 0
3 years ago
1]
balu736 [363]

Explanation:

Given -

  • An organic compound gives H₂ gas with Na
  • On treatment with alkaline iodine it gives yellow ppt.
  • On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)

To Find -

  • Name the compound and write the reaction involved

Now,

Let A be the organic compound.

Then,

  1. A + Na → + H₂↑
  2. A + I₂ → CHI₃ (yellow ppt.)
  3. A + CrO₃ + H⁺ → C₂H₄O

Now,

Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.

  • Functional group of aldehyde = —CHO

And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).

Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.

It means,

We know that 1° alcohol on oxidation gives aldehyde.

Here it gives 2 Carbon aldehyde.

It means,

Here 2 Carbon and 1° alcohol is used.

Now,

Its cleared that Compound A is Ethanol.

Reaction Involved -

  1. CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
  2. CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
  3. CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO

6 0
3 years ago
Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
3 years ago
According to the ideal gas law, increasing the volume of a closed reaction container decreases the thermal energy because the
hichkok12 [17]

Answer:

b is ur answer the temputer does increase

Explanation:

8 0
3 years ago
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