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3 years ago

18°C = _____ -255 K 0 K 18 K 291 K

Chemistry

2 answers:

horrorfan [7]3 years ago

7 0

T K = ºC + 273

T = 18 + 273

T = 291 K

hope this helps!

g100num [7]3 years ago

3 0

Answer:

T = 291 K

Explanation:

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On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of

Murrr4er [49]

Answer:

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Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

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Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of $SO_2$ gas at STP occupying 1.0 L = n''

$n''\times 22.4L=1.0L$

$n''=\frac{1.0 L}{22.4 L}=0.04464 mol$

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

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Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = $\frac{0.08920 mol}{0.04464 mol}=2$

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Sulfur = $\frac{0.04464 mol}{0.04464 mol}=1$

The empirical formula of the organic compound is = $C_2H_6S_1$

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3 years ago

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3 years ago

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3 years ago

A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet

stira [4]

Explanation:

The given data is as follows.

Mass of apple sauce mixture = 454 kg

Heat added (Q) = 121300 kJ

Heat capacity ( $C_{p}$ ) of apple sauce at $32.8^{o}C$ = 4.0177 $kJ/kg^{o}C$

So, Heat given by heat exchanger = heat taken by apple sauce

Q = $mC_{p} \Delta T$

or, Q = $mC_{p} (T_{f} - T_{i})$

Putting the given values into the above formula as follows.

Q = $mC_{p} (T_{f} - T_{i})$

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Thus, we can conclude that outlet temperature of the apple sauce is $76.5^{o}C$ .

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3 years ago