Explain how Earth is heated through radiation.

We are asked to convert 25 cg to units of hg.

1 cg = 1 centigram = 10⁻² g
1 hg = 1 hectogram = 10² g

The options given are:

a) 1 hg/ 10² g
b) 10² cg/ 1 hg
c) 10² hg/ 1 cg
d) 10⁻² g/ 1 cg

To convert 25 cg to 1 hg, we could convert the 25 cg to grams first, then grams to hg.

25 cg · 10⁻² g/ 1cg = 0.25 g

Here we have converted our number from cg to grams. We can use another conversion of grams to hg to complete the conversion.

0.25 g · 1 hg/ 10² g = 0.0025 hg

Therefore, the first conversion we used was d) 10⁻² g/ 1 cg.

4 0

3 years ago

A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this

NeTakaya

Answer:

a) molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane $CH_4}= 16 g/mol$

Molar mass of ethane $C_2H_6= 30 g/mol$

Molar mass of ethylene $C_2H_4 = 28 g/mol$

Molar mass of water $H_2O=18g/mol$

number of moles = $\frac{mass}{molar mass}$

Their molar composition can be calculated as follows:

$n_{CH_4}= \frac{75}{16}$

$n_{CH_4}=$ 4.69 moles

$n_{C_2H_6} = \frac{10}{30}$

$n_{C_2H_6} =$ 0.33 moles

$n_{C_2H_4} = \frac{5}{28}$

$n_{C_2H_4} =$ 0.18 moles

$n_{H_2O}= \frac{10}{18}$

$n_{H_2O}=$ 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = $\frac{n_{H_2O}}{n_{drygas}}$

= $\frac{0.56}{5.2}$

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

$CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O$

4.69 2× 4.69

moles moles

$C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O$

0.33 3.5 × 0.33

moles moles

$C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O$

0.18 3× 0.18

moles moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

$CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h$

Total moles of $O_2$ required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles $O_2$

Thus, moles of air required = $\frac{1}{0.21}*11.075$

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0

4 years ago