All categories

4 years ago

Explain how heat would change the density of a parcel of air

1 answer:

6 0

<span>Heat would make the molecules move faster so they would spread out therefore making air less dense.
Less dense means less packed. Heat increase the kinetic energy</span>

You might be interested in

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.

padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

#SPJ4

5 0

1 year ago

Do electric or magnetic fields affect the x-rays?

Levart [38]

Electric and magnetic fields do not affect xrays as they only affect charged particles and xrays have no charge

hope that helps

3 0

3 years ago

What is the freezing point in °C) of a 0.743 m

lesantik [10]

Answer:

-0.276 Degrees C

Explanation:

kf of water is 1.86

Freezing point depression= m x kf x i

i= ions present ( K+ Cl-) 1=2

1.86x2X.743= .276

Since its freezinf point depression the freezing point will lower

0-.276= -.276

7 0

3 years ago

What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium c

LuckyWell [14K]

Answer:

[A g + ] = 3.12 *10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^− 12

Step 2: The balanced equation

2 A g + ( a q ) + C rO4^2- ( a q ) →Ag2CrO4 ( s)

Step 3:

[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 -0.088 = 0.123 M

1.2 * 10^− 12 = ( 2 x ) ²*( 0.123M + x )

[ C O 3^ −2] f >> x

1.2 * 10^− 12 = ( 2 x ) ²* 0.123M

x = 1.56 * 10^-6

[A g + ] f = 2x = 3.12 *10^-6 M

4 0

3 years ago

What, according to you, are the necessary activities that a penetration tester needs to perform while conducting a penetration t

bogdanovich [222]

Answer: First of all planning is done to set the goals to how to test the target. then scanning is done. after that target is uncovered to see how the target exploits by the use of web application attacks. then analysis is done an tests of the penetration tests are compiled.

Explanation:

6 0

3 years ago