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murzikaleks [220]
3 years ago
9

What is the partial pressure of C? atm C

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
8 0

The partial pressure of carbon is 45 mm Hg.

Explanation:

  • The partial pressure of carbon dioxide is referred as the amount of carbon dioxide present in venous or arterial blood. It acts as a ventilation in the lungs.
  • There is a formula for measuring partial pressure . As we know total pressure means summation of the pressure of all the gases included .
  • To find partial pressure we need- total pressure* fraction of mole of that gas. The partial pressure of CO2 is more because it carries deoxygenated blood from the whole body towards the lungs.

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If the elastic modulus of cobalt (co) is 200 gpa, and the elastic modulus of tungsten carbide (wc) is 700 gpa, calculate the upp
Dmitry_Shevchenko [17]
To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp
7 0
4 years ago
PLEASE HELP!!!
Kaylis [27]
I think it comes from Fungi. I'm not 100% sure about that though.
8 0
3 years ago
What is the mass of 7 x 10^28 atoms of Fe?
Nostrana [21]

Answer:

6 x 10⁶ g Fe

Explanation:

Step 1: Set up dimensional analysis

7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)

Step 2: Multiply, divide, and cancel out units

atoms Fe and atoms Fe cancel out.

mol Fe and mol Fe cancel out.

We should be left with g Fe.

7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe

116279(55.85) = 6.49 x 10⁶ g Fe

Step 3: Sig figs

There is only 1 sig fig in this problem.

6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe

4 0
3 years ago
A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
4 years ago
Chlorination of ethane yields, in addition to ethyl chloride, a mixture of two isomeric dichlorides. what are the structures of
jeyben [28]
CH_3 CH_2 Cl has two different types of hydrogen. On further chlorination they are being substituted by chlorine to give two isomeric dichlorides

The structures of the two chlorides are shown below.

Where Hb is replaced by Cl and Ha is replaced by Cl

5 0
4 years ago
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