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S_A_V [24]
3 years ago
7

Calculate how much Carbon Dioxide IN GRAMS needs to be absorbed to

Chemistry
1 answer:
Alja [10]3 years ago
4 0

Answer:

687.5 g of CO2

Explanation:

We'll begin by writing the balanced equation for photosynthesis. This is illustrated below:

6CO2 + 6H2O —> C6H12O6 + 6O2

Next, we shall determine the mass of CO2 absorbed and the mass of O2 released from the balanced equation. This is illustrated below:

Mass of CO2 = 12 + (2 × 16) = 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 6 × 44 = 264 g

Molar mass of O2 = 2 × 16 = 32 g/mol

Mass of O2 from the balanced equation = 6 × 32 = 192 g

Summary:

From the balanced equation above,

264 g of CO2 were absorbed to release 192 g of O2.

Finally, we shall determine the mass of CO2 absorbed to release 500 g of O2. This can be obtained as follow:

From the balanced equation above,

264 g of CO2 were absorbed to release 192 g of O2.

Therefore, Xg of CO2 will be absorbed to release 500 g of O2 i.e

Xg of CO2 = (264 × 500)/192

Xg of CO2 = 687.5 g

Thus, 687.5 g of CO2 will be absorbed to release 500 g of O2.

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Butane, C4H10 has a standard enthalpy of combustion= -2881 kJ/mol
Bad White [126]

Answer:

14.91 K.

Explanation:

  • To solve this problem, we can use the following relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat transferred to water.

m is the mass of the amount of water (m = 2.0 kg = 2000.0 g).

c is the specific heat capacity of water (c = 4.2 J/g.K).

ΔT is the change in temperature due to the transfer of butane burning.

  • To determine Q that to be used in calculation:

Q from 4.000 g of butane is completely burned is - 198.3 kJ = 198300 J.

<em>The negative sign</em><em> symbolizes the the enthalpy change is </em><em>exothermic</em><em>, which means that </em><em>the</em><em> </em><em>energy is released</em><em>. </em>

  • Note that only 63.15% of the energy generated is actually transferred to the water.

∴ Q (the amount of heat transferred to water) = (198300 J)(0.6315) = 125226.45 J.

  • Now, we can obtain the change in temperature:

∴ ΔT = Q/m.c. = (125226.45 J) / (2000.0 g)(4.2 J/g.K) = 14.9079 K ≅ 14.91 K.

<em>This means that the temperature is increased by 14.91 K.</em>

<em />

5 0
3 years ago
Help I will give you 15 points help now I MEAN RIGHT NOW PLZ
nataly862011 [7]
1. A

2. D.

3. Either A or B (my best guess is B.)
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4 years ago
What ionic are compounds used in our daily life?
Gemiola [76]

Answer:

Explanation:

NaCl sodium Chloride as ordinary salt

Naf sodium fluoride is used in toothpaste which we use every day

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3 years ago
Describe the best method for separating a mixture of sand, salt, and steel shavings by placing the steps below in the correct or
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Answer: a sifter

Explanation:

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3 years ago
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Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)
vichka [17]

Answer:

Kb →  1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m =  Kb →  1.56 °C / m

3 0
4 years ago
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