What biological processes can add co2 to the atmosphere?

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

3 years ago

In the spring of 1984, concern arose over the presence of ethylene dibromide, or EDB, in grains and cereals. EDB has the molecul

Lady bird [3.3K]

869.6 × 10¹⁴ molecules of EDB

Explanation:

We have 1.9 lb of flour with a EDB concentration of 31.5 ppb.

We need to transform lb in grams.

1 lb = 453.6 grams

1.9 lb = (1.9 × 453.6) / 1 = 861.8 grams

Now we determine the number of molecules of EDB in the sample by devise the following reasoning:

if we have 31.5 × 10⁻⁹ g of EDB in 1 g of sample

then we have X g of EDB in 861.8 g of sample

X = (31.5 × 10⁻⁹ × 861.8) / 1 = 27146.7 × 10⁻⁹ g of EDB

Molecular mass of EDB (C₂H₄Br₂) = 188 g/mole

Taking in account that 1 mole of any substance contains 6.022 × 10²³ (Avogadro’s number) molecules we devise the following reasoning:

if 188 g of EDB contains 6.022 × 10²³ molecules

then 27146.7 × 10⁻⁹ g of EDB contains Y molecules

Y = (27146.7 × 10⁻⁹ × 6.022 × 10²³) / 188 = 869.6 × 10¹⁴ molecules of EDB

Learn more:

about Avogadro’s number

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8 0

4 years ago

Read 2 more answers

Assuming that a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, determine how many kg of co2 are prod

sleet_krkn [62]

Answer: -

If a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, 0.26 kg of CO₂ are produced for each tank of gasoline burned.

Explanation: -

Density of the gasoline = 0.77 kg / liter

Volume of the tank containing the gasoline = 80 liter.

Mass of gasoline produced from each tank

= Volume of the tank containing the gasoline x Density of the gasoline

= $\frac{0.77 kg}{1 liter}$ x 80 liter

= 61.6 kg

Chemical formula of gasoline = C₈H₁₈

Molar mass of gasoline C₈H₁₈ = 12 x 8 + 1 x 18 = 114 g/ mol

Number of moles of C₈H₁₈ = $\frac{61.6 g}{114 g}$ x 1 mol

= 0.54 mol of C₈H₁₈

The chemical equation for the burning of gasoline is

2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

From the balanced equation we see

2 mol of C₈H₁₈ gives 16 mol of CO₂

0.54 mol of C₈H₁₈ gives $\frac{16 mol CO2 x 0.54 mol C8H18}{2 mol C8H18}$ mol of CO₂

= 4.32 mol of CO₂

Molar mass of CO₂ = 12 x 1 + 16 x 3 =60 g / mol

Mass of CO₂ = Molar mass of CO₂ x Number of moles of CO₂

= $\frac{60g x 4.32 mol}{1 mol}$

= 259.2 g

= $\frac{259.2}{1000}$

= 0.259 Kg

= 0.26 kg rounded off to 2 significant figures.

Thus if a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, 0.26 kg of CO₂ are produced for each tank of gasoline burned.

4 0

3 years ago

Cis-4-tertButylcyclohexyl bromide (compound 1) and Trans-4 tert Butylcyclohexylbromide (compound 2) are reacted with Potassium T

Mkey [24]

Answer:

In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.

Explanation:

In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.

Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.

This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.

4 0

3 years ago

A compound is 54.53% c, 9.15% h, and 36.32% o by mass. what is its empirical formula? the molecular mass of the compound is 132

Yakvenalex [24]

<span>First - you need the empirical formula.

So, assume you have 100 g of the compound.

If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.

54.53 g C (1 mole C / 12.01 g C) = 4.540

9.15 g H (1 mole H / 1.008 g H) = 9.077

36.32 g O (1 mole O / 15.9994 g O) = 2.270

Take the smallest number found and divide the others by it to get the empirical formula.

4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.

So, that gives you the empirical formula of C2H4O.

Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.

132/44 = 3.

So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>

6 0

3 years ago