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3 years ago

C2H5OH(g) + 302(g) → 2CO2(g) + 3H2O(g) + heat reaction type

Chemistry

1 answer:

Tcecarenko [31]3 years ago

5 0

Answer:

Combustion

Explanation:

The reaction type is a combustion reaction;

Ethanol + oxygen gas → carbon dioxide + water

Now;

Ethanol is the fuel and oxygen gas rapidly combines with it to form carbon dioxide and water.

This process releases heat energy from chemical bonds.
Most hydrocarbons burn in excess air or oxygen to give carbon dioxide and water.

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In an acid/base lab, a student used blue litmus paper and placed it in the sample. It stayed blue. Then the student used red lit

Allushta [10]

Answer: The substance is neutral

Explanation:

If you place red litmus paper into a basic substance, it turn's blue. When it comes in contact with an acidic or neutral substance, it doesn't change colour it remains red.

Blue litmus paper in acid turns red when placed in a basic or neutral solution it remains blue. From this it can be concluded that the solution is neutral.

7 0

3 years ago

C5H12 + 8O2 → 5CO2 + 6H2O Calculate the energy change for the reaction:

KATRIN_1 [288]

C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol

C (s) + O2 (g) -----> CO2 (g)
Delta H = -393.5 kJ/mol

H2 (g) + (1/2)O2 (g) ------> H2O (l)
Delta H = -286 kJ/mol

Possible answers:
a. +35 kJ/mol
b. + 1,073 kJ/mol
c. -4,185 kJ/mol
d. -2,826 kJ/mol
e. -178 kJ/mol

8 0

3 years ago

A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o

Ludmilka [50]

Answer:

0.297 °C

Step-by-step explanation:

The formula for the freezing point depression ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles i = 1

Data:

Mass of glucose = 10.20 g

Mass of water = 355 g

ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

(a) Moles of glucose

n = 10.20 g × (1 mol/180.16 g)

= 0.056 62 mol

(b) Kilograms of water

m = 355 g × (1 kg/1000 g)

= 0.355 kg

(c) Molal concentration

b = moles of solute/kilograms of solvent

= 0.056 62 mol/0.355 kg

= 0.1595 mol·kg⁻¹

(d) Freezing point depression

ΔT_f = 1 × 1.86 × 0.1595

= 0.297 °C

3 0

3 years ago

A little boy has placed his pet squirrel into a balloon with a volume of 3L at a pressure of 1.0 atm. If the boy takes his pet s

s2008m [1.1K]

Answer: The squirrel's balloon will be 0.86 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=1.0atm\\V_1=3L\\P_2=3.5atm\\V_2=?$

Putting values in above equation, we get:

$1.0\times 3L=3.5\times V_2\\\\V_2=0.86$

Thus the squirrel's balloon will be 0.86 L

7 0

3 years ago

Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an

Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0

3 years ago