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Valentin [98]
3 years ago
7

In a car piston shown above, the pressure of the compressed gas (red) is 5.00 atm. If the area of the piston is 0.0760 m^2. What

is the force exerted on the piston in Newtons (N)?
Chemistry
1 answer:
Shkiper50 [21]3 years ago
6 0

Answer:

38503.5N

Explanation:

Data obtained from the question include:

P (pressure) = 5.00 atm

Now, we need to convert 5atm to a number in N/m2 in order to obtain the desired result of force in Newton (N). This is illustrated below:

1 atm = 101325N/m2

5 atm = 5 x 101325 = 506625N/m^2

A (area of piston) = 0.0760 m^2

Pressure is force per unit area. Mathematically it is written as

P = F/A

F = P x A

F = 506625 x 0.0760

F = 38503.5N

Therefore, the force exerted on the piston is 38503.5N

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What did Ernest Rutherford s gold foil experiment demonstrate about atoms?
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A gas has density 2.41 g/liter at 25°C and 770 mm Hg. Calculate it's molecular mass (R = 0.0821 L atm.mol-1K-1 ​
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Answer:

Explanation:

Given : Density - 2.41 g/liter  

Temperature - 25° C  

Pressure : 770 mm Hg  

R = 0.0821 L atm mol-¹K-¹

 

Find : Molecular mass of gas

Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.

Keeping the values in equation-  

Pressure : 770 mm Hg = 1 atm  

Temperature : 273 + 25 = 298 K

 

M = dRT/P  

M = (2.41*0.0821*298)/1  

M = 58.96 gram/mol

Thus, the molecular mass of gas is 58.96 gram/mol.

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35. In the collision theory, a collision that leads to the formation of products is called an
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Hope my answer has helped you!

7 0
3 years ago
Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

8 0
3 years ago
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