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3 years ago

_P4 + __O2 → P2O3 How do I balance this equation?

Chemistry

2 answers:

enot [183]3 years ago

7 0

Explanation:

hope it's helpful for you

pls mark above guy ans as brainliest

Murljashka [212]3 years ago

5 0

Explanation:

please mark me as brainlest

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C6H10 Hope I helped! :P

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3 years ago

What element has greatest atomic radius from group 3?

stira [4]

Answer:

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Explanation:

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3 years ago

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,

A + Na → + H₂↑
A + I₂ → CHI₃ (yellow ppt.)
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 -

CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
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

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

_P4 + __O2 → P2O3 How do I balance this equation?