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Rzqust [24]
3 years ago
7

How many grams of ethanol, C2H5OH, can be boiled with 843.2 kJ of heat energy? The molar heat of vaporization of ethanol is 38.6

kJ/mol.
Chemistry
1 answer:
Katarina [22]3 years ago
7 0

Answer:

1.00 × 10³ g

Explanation:

Step 1: Given data

  • Provided heat (Q): 843.2 kJ
  • Molar heat of vaporization of ethanol (ΔH°vap): 38.6 kJ/mol

Step 2: Calculate the moles of ethanol vaporized

Vaporization is the passage of a substance from liquid to gas. We can calculate the number of moles (n) vaporized using the following expression.

Q = ΔH°vap × n

n = Q / ΔH°vap

n = 843.2 kJ / (38.6 kJ/mol) = 21.8 mol

Step 3: Calculate the mass corresponding to 21.8 moles of ethanol

The molar mass of ethanol is 46.07 g/mol.

21.8 mol × 46.07 g/mol = 1.00 × 10³ g

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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
What weight of sodium hydroxide will react with 73 gram of hydrogen chloride gas at NTP to Produce 117.0 gram of Nacl and 36 gra
Wewaii [24]

Let's write the equation

\\ \sf\longmapsto {NaOH\atop ?}+{HCl\atop 73g}\longrightarrow {NaCl\atop 117g}+{H_2O\atop 36g}

According to law of conservation of mass .

  • Mass of products=Mass of reactants

Let required value be x

\\ \sf\longmapsto x+73=117+36

\\ \sf\longmapsto x+73=153

\\ \sf\longmapsto x=153-73

\\ \sf\longmapsto x=80g

8 0
2 years ago
Read 2 more answers
HELP PLZ ITS A TEST.
tensa zangetsu [6.8K]

Answer:

The answer to your question is below

Explanation:

1 ) Which of the following is a chemical property.

A. Ability to become magnetic  this is wrong, it's a physical property

B. Ability to conduct electricity   this is wrong, it's a physical property

C. Ability to conduct heat   this is wrong, it's a physical property

D. Ability to react with acid This is the right answer

2 ) You heat two substances, A and B. Both substances change color. When cooled, substance B returns to its original color, but substance A does not. What most likely happened?.

A. Physical change occurred in both substances   This is wrong, a physial change only happens in substance B

B. A physical change occurred in substance A, and a chemical change occured in substance B.  This is wrong, a physical change occurred in substance B and a chemical change occurred in substance A.

C. A chemical change occurred in substance A, and a physical change occurred in substance B.  This is the right answer is exactly what happens

D. A chemical change occurred in both substances  This answer is wrong.

3 ) how many grams are in 2.5 pound sample? Express your answer in scientific notation with three significant figures. ( 1 pound= 453.592 grams).

Rule of three

    1 pound ------------------- 453.592 g

  2.5 pounds ---------------   x

  x = 2.5(453.592)/1

  x = 1133.98 g = 1.13 x 10³ g    letter D

7 0
3 years ago
Read 2 more answers
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

a)

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

b)

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
The solubility of N2 in blood can be a serious problem (the "bends") for divers breathing compressed air (78% N2 by volume) at d
OLga [1]

Answer:

The volume is 19.7 mL

Explanation:

<u>Step 1</u>: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

<u>Step 2</u>: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.

13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L

Step 3: Calculate volume

P*V=nRT

with P = 1.00 atm

with V = TO BE DETERMINED

with n =  8.04*10^-4 mol/L  *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml

The volume is 19.7 mL

5 0
3 years ago
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