Which type of microscope can be used to view cellular organelles such as the endoplasmic reticulum and Golgi?

pentagon [3]

<span>
It contains 6.02 mc001-1.jpg 1023 grams of sodium chloride.

It is the mass of 12 carbon atoms.

It contains 6.02 mc001-2.jpg 1023 particles of a given substance.</span>

5 0

3 years ago

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The mass of a single uranium atom is 4.70x10^-22 grams. How many uranium atoms would there be in 111 milligrams of uranium?

emmainna [20.7K]

2.4 x 10²² atoms

<h3>Further explanation</h3>

Atomic mass is the average atomic mass of all its isotopes

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu

So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom

mass single Uranium atom=4.7 x 10⁻²² g

then for 111 mg=0.111 g

$\tt \dfrac{0.111}{4.7\times 10^{-22}}=2.4\times 10^{22}$

5 0

2 years ago

For test 6 (150 hevy 50 light particles), what is the pressure contribution from the heavy particles?

lawyer [7]

Answer:

The pressure contribution from the heavy particles is 17.5 atm

Explanation:

According to Dalton's law of partial pressures, if there is a mixture of gases which do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture.

In the simulation:

the pressure of the 50 light particles alone was determined to be 5.9 atm, the pressure of the 150 heavy particles alone was measured to be 17.5 atm,

the total pressure of the mixture of 150 heavy and 50 light particles was measured to be 23.4 atm

Total pressure = partial pressure of Heavy particles + partial pressure of light particles

23.4 atm = partial pressure of Heavy particles + 5.9 atm

Partial pressure of Heavy particles = (23.4 - 5.9) atm

Partial pressure of Heavy particles = 17.5 atm

Therefore, the pressure contribution from the heavy particles is 17.5 atm

4 0

2 years ago

Which element is best known for it's bad smell?

tia_tia [17]

The bad smelling element is Sulfur, which smells like rotting eggs.

5 0

2 years ago

A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L

slega [8]

Answer:

5.41 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

$Moles=0.033 \times {0.45}\ moles$

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.01485\ g= \frac{Mass}{342.39\ g/mol}$

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

<u>Answer - 5.41 g</u>

5 0

3 years ago