A layer of sandstone is found on top of a layer of shale

How much work (in J) is involved in a chemical reaction if the volume decreases from 4.35 to 1.20 L against a constant pressure

Marianna [84]

Answer:

W = -262 J.

Explanation:

Hello there!

In this case, according to the given information, we can recall the definition of work in terms of constant pressure and variable volume as follows:

$W=P(V_2-V_1)$

So we plug in the given pressure and volumes to obtain:

$W=0.822atm(1.20L-4.35L)\\\\W=-2.60atm*L$

Now, we convert this number to J (Pa*m³) by using the shown below conversion factor:

$W=-2.60atm*L*\frac{101325Pa}{1atm} *\frac{1m^3}{1000L}\\\\W=-262J$

Regards!

3 0

3 years ago

CaC2(s) + 2H2O(l) --> Ca(OH)2(aq) + C2H2(g) In the reaction above, 0.5487 grams of calcium carbide are completely consumed to

jasenka [17]

Answer:

239.7mL

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

The balanced chemical equation in this question is as follows:

CaC2(s) + 2H2O(l) --> Ca(OH)2(aq) + C2H2(g)

From the equation, 1 mole of CaC2 produces 1 mole of ethylene gas, C2H2.

Using mole = mass/molar mass

Molar mass of CaC2 = 40 + 12(2)

= 40 + 24

= 64g/mol

mole = 0.5487/64

mole = 0.00857mol of CaC2

Hence, 0.00857mol of CaC2 produced 0.00857mol of C2H2

Based on the information provided, n = 0.00857mol, T = 43°C = 43 + 273 = 316K, p = 0.926 atm

PV = nRT

V = nRT/P

V = 0.00857 × 0.0821 × 316/0.926

V = 0.222/0.926

V = 0.2397L

In mL, volume = 0.2397 × 1000

= 239.7mL

3 0

3 years ago

How did the industrial revolution contribute to global climate change

Deffense [45]

Answer:humans have put ridiculous amounts of carbon dioxide into the atmosphereThese events are linked to the mass burning of fossil fuels to meet an increase in human demand

So the answer is True

Explanation:

6 0

2 years ago

For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →

Vinvika [58]

Answer : The oxidizing element is N and reducing element is O.

$KNO_{3}$ is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

$KNO_{3}$ act as an oxidizing agent.

The oxidation number of N in $KNO_{3}$ is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in $KNO_{2}$ is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number reduced this means $KNO_{3}$ itself get reduced to $KNO_{2}$ and it can act as an oxidizing agent.

$KNO_{3}$ act as a reducing agent.

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

The oxidation number of O in $KNO_{3}$ is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in $O_{2}$ is Zero (o).