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3 years ago

Vitreous humor is located wear?

Chemistry

1 answer:

Vikentia [17]3 years ago

3 0

Eye

Explanation:

Vitreous humor is found in the human eye and other animals.

It is a gel that fills the space between the lens and retina of the eye. This matter helps to keep the shape of the eye in place by maintaining a constant pressure in the eye.

It is typically made up of water, gelatinous and transparent.

Learn more:

Human eye brainly.com/question/8032392

Color in the eye brainly.com/question/9434044

#learnwithBrainly

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A bottle of wine contains 0.21 moles of ethanol,C2H5OH. The volume of the solution is 0.1 L. Calculate the molarity.

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Answer:

2.1mol/L

Explanation:

Number of moles = 0.21 moles

Volume = 0.1L

Molarity of a substance is the number of moles of solute dissolved in a volume of solvent (L)

Molarity = number of moles / volume of solvent

Molarity = 0.21 / 0.1

Molarity = 2.1mol/L

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A bobsled has a momentum of 4000 0 kg* m/s to the south. Friction on the track reduces its momentum to 500 kg* m/s to the south.

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3 years ago

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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

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3 years ago

At a birthday pool party, the temperature is 28.50°C and the atmospheric pressure is 755.4 mmHg. One of the decoration helium ba

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Answer:

The volume of the balloon will be 5.11L

Explanation:

An excersise to solve with the Ideal Gases Law

First of all, let's convert the pressure in mmHg to atm

1 atm = 760 mmHg

760 mmHg ___ 1 atm

755.4 mmHg ____ (755.4 / 760) = 0.993 atm

922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm

T° in K = 273 + °C

28.5 °C +273 = 301.5K

26.35°C + 273= 299.35K

P . V = n . R .T

First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K

(0.993atm . 6.25L) / 0.082 . 301.5 = n

0.251 moles = n

Second situation:

1.214 atm . V = 0.251 moles . 0.082 . 301.5K

V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm

V = 5.11L

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Vitreous humor is located wear?​

Vitreous humor is located wear?