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Vladimir [108]
3 years ago
14

An organelle that is not found in this illustration of a cell would be

Chemistry
2 answers:
zloy xaker [14]3 years ago
8 0

Answer:

The answer is cell wall

Explanation:

Because it is

nekit [7.7K]3 years ago
3 0
It would be A. Cell wall
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Vitreous humor is located wear?​
Vikentia [17]

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

3 0
2 years ago
How many orbitals surround the nucleus in a neutral atom of sulfur (S)?
Leto [7]
<h2>C</h2>

Explanation:

The atomic number of S is 16

So,number of electrons in S is 16

The electronic configuration of S is 2,8,6

The orbital electronic configuration of S is ^{1}s_{2}^\text{ }^{2}s_{2}^\text{ }^{2}p_{x2}^\text{ }^{2}p_{y2}^\text{ }^{2}p_{z2}^\text{ }^{3}s_{2}^\text{ }^{3}p_{x2}^\text{ }^{3}p_{y1}^\text{ }^{3}p_{z1}^\text{ }

So,the number of orbitals involved is 9.

3 0
3 years ago
A newly discovered element, Z, has two naturally occurring isotopes. 90.3 percent of the sample is an isotope with a mass of 267
blsea [12.9K]
atomic mass=percentage of isotope a * mass of  isotope a + percentage of isotope b * mass of  isotope b+...+percentage of isotope n * mass of isotope n.

Data:
mass of isotope₁=267.8 u
percentage of isotope₁=90.3%

mass of isotope₂=270.9 u
percentage of isotope₂=9.7%

Therefore:

atomic mass=(0.903)(267.8 u)+(0.097)(270.9 u)=
=241.8234 u + 26.2773 u≈268.1 u

Answer: the mass atomic of this element would be 268.1 u
7 0
3 years ago
1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th
vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy,  , for a homogeneous substance to be a function of specific volume  and temperature  .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T)  dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0
2 years ago
How much is the oxygen valency?
Marina CMI [18]
The oxygen valency is 2
4 0
2 years ago
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