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3 years ago

Please help me will reward Brainly

Chemistry

2 answers:

Damm [24]3 years ago

5 0

Is it carbon dioxide

zhannawk [14.2K]3 years ago

5 0

The answer is Carbon Dioxide (Co2).

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The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

2 years ago

Why do you use 6.02*10^23 here?? #21???

almond37 [142]

6.02*10^23 is Avagadro's number representing the number of molecules per mole of substance.

6 0

2 years ago

Read 2 more answers

In the following compound (HCl) How many electrons are gained and lost by each atom?

Verizon [17]

Explanation:

In HCL, one positive atom is given to chlorine from hydrogen so that it can complete it's octate. chlorine take one electron from hydrogen.

In NaCl, Sodium takes one electron from chlorine to complete its orbit with eight electrons. Chlorine gives one electron to sodium.

8 0

3 years ago

What is the most important reason to publish results as part of the scientific process?

svp [43]

<span>It allows other scientists to check findings</span> is the most important reason to publish results as part of the scientific process.

4 0

2 years ago

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A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

$NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)$

So, $OH^{-}$ is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of $OH^{-}$ increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant.

Therefore excess amount of $OH^{-}$ combines with $NH_{4}^{+}$ to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0

3 years ago