<span>Table salt is inorganic TNT is organic Glucose is organic 2,4-D is organic Limestone is inorganic Water is inorganic What makes a compound organic is the presence of a carbon, with the exception of cabonates. In this case all of the compounds in this list that have carbon except for CaCO3, are organic and the other compounds are inorganic.</span>

3 0

4 years ago

Silicon carbide (SiC) is an important ceramic material made by reacting sand (silicon dioxide, SiO2) with powdered carbon at a h

makkiz [27]

Answer:

Percent yield of SiC is 77.0%.

Explanation:

Balanced reaction: $SiO_{2}+3C\rightarrow SiC+2CO$

Molar mass of SiC = 40.11 g/mol

Molar mass of $SiO_{2}$ = 60.08 g/mol

So, 100.0 kg of $SiO_{2}$ = $\frac{100.0\times 10^{3}}{60.08}$ moles of $SiO_{2}$ = 1664 moles of $SiO_{2}$

According to balanced equation, 1 mol of $SiO_{2}$ produces 1 mol of SiC

Therefore, 1664 moles of $SiO_{2}$ produce 1664 moles of SiC

Mass of 1664 moles of SiC = $(1664\times 40.11)g$ = 66743g = 66.74 kg (4 sig. fig.)

Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)] $\times 100$ %

= $\frac{51.4kg}{66.74kg}\times 100$ %

= 77.0%

4 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago

FREÉ Points out!!!!!!!!!!!​

FREÉ Points out!!!!!!!!!!!