1. In which order does food travel through the digestive

Look at the wave in the picture. If it has a wavelength of 2 cm, how far apart are points A and B? How do you know?

Harlamova29_29 [7]

Answer:

picture is not there.....

8 0

3 years ago

Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper

KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0

3 years ago

If 36 000 kg of full cream milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream

Yuri [45]

Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

-XgA = percentage of fat in A = 0.04

We define output variables.

- skim milk (B)

-creme (C)

-XgB = fat percentage at B = 0.0045

-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

Input = Output

Balance sheet

1) A = B + C

Fat balance

2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values of the outflows.

C=478.11kg/h

B=5522.33kg/h

7 0

3 years ago

Liquids and gases are matter because they take up space and have

Aneli [31]

Volume. Gases and liquids are typically measured in milliliters (mL) or cubic centimeters (cm^3) - both of which are equivalent (1 mL = 1 cm^3).

8 0

3 years ago

A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g of chlorine pentafluoride gas. You can assume both

Jobisdone [24]

Answer:

Total pressure: 2.89 atm

Mole fraction CO: 0.88

Partial pressure CO: 2.56 atm

Mole fraction ClF₅: 0.12

Partial pressure ClF₅: 0.33 atm

Explanation:

We should apply the Ideal Gases Law to solve this:

P . V = n . R . T

We need n, which is the total moles for the mixture

Total moles = Moles of CO + Moles of ClF₅

Moles of CO = mass of CO / molar mass CO → 18 g/28 g/mol = 0.643 mol

Moles of ClF₅ = mass of ClF₅ / molar mass ClF₅ → 10.6g/ 130.45 g/m = 0.0812 mol

0.643 mol + 0.0812 mol → 0.724 moles in the mixture

So we have the total moles so with the formula we would know the total pressure.

P . 6L = 0.724 mol . 0.082L.atm/mol.K . 292.2K

P = ( 0.724 mol . 0.082L.atm/mol.K . 292.2K) / 6L

P = 2.89 atm

Mole fraction is defined as the quotient between the moles of gas over total moles, and it is equal to partial pressure of that gas over total pressure

Moles of gas X /Total moles = Partial pressure of gas X/Total pressure

(Moles of gas X / Total moles) . Total pressure = Partial pressure of gas X

Mole fraction CO = 0.643 / 0.724 = 0.88

Partial pressure CO = 0.88 . 2.89 atm → 2.56 atm

Mole fraction ClF₅ = 0.0812 / 0.724 = 0.12

Partial pressure ClF₅ = 0.12 . 2.89 atm → 0.33 atm

5 0

3 years ago