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3 years ago

PLEASE HELP!!!! :D When you look at the Sun through a filtered telescope, the visible portion of the Sun appears blotchy.

2 answers:

5 0

<span>When an individual looks through a filtered telescope in which he or she observes the sun, the portion where it appears blotchy is likely to be called the sunspots while the layer of the sun where it shows where it occurs is called the photosphere.</span>

olga2289 [7]3 years ago

4 0

The answer is; sunspots, photosphere.

Sunspots are cooler than the rest of the surface of the sun hence their appearance of being darker than the rest of the photosphere. The photosphere is the outer sun’s shell that emits radiation to space. These sunspots are associated with increased magnetic influx that is accompanied by coronal mass ejection. It is thought that these magnetic fluxes hinder efficient convection currents in the regions.

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3 years ago

A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density

Yuliya22 [10]

The solution of Sulfuric Acid (H2SO4) has the following mole fractions:

mole fraction (H2SO4)= 0.034
mole fraction (H2O)= 0.966

To solve this problem the formula and the procedure that we have to use is:

n = m / MW
= ∑ AWT
mole fraction = moles of A component / total moles of solution
ρ = m /v

Where:

m = mass
n = moles
MW = molecular weight
AWT = atomic weight
ρ = density
v = volume

Information about the problem:

m solute (H2SO4) = 17.75 g
v(solution) = 100 ml
ρ (solution)= 1.094 g/ml
AWT (H)= 1 g/mol
AWT (S) = 32 g/mol
AWT (O)= 16 g/mol
mole fraction(H2SO4) = ?
mole fraction(H2O) = ?

We calculate the moles of the H2SO4 and of the H2O from the Pm:

MW = ∑ AWT

MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4

MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)

MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol

MW (H2SO4)= 98 g/mol

MW (H2O)= AWT (H) * 2 + AWT (O)

MW (H2O)= (1 g/mol * 2) + (16 g/mol)

MW (H2O)= 2 g/mol + 16 g/mol

MW (H2O)= 18 g/mol

Having the Pm we calculate the moles of H2SO4:

n = m / MW

n(H2SO4) = m(H2SO4) / MW (H2SO4)

n(H2SO4) = 17.75 g / 98 g/mol

n(H2SO4) = 0.1811 mol

With the density and the volume of the solution we get the mass:

ρ(solution)= m(solution) /v(solution)

m(solution) = v(solution) * ρ(solution)

m(solution) = 100 ml * 1.094 g/ml

m(solution) = 109.4 g

Having the mass of the solution we calculate the mass of the water in the solution:

m(H2O) = m(solution) - m solute (H2SO4)

m(H2O) = 109.4 g - 17.75 g

m(H2O) = 91.65 g

We calculate the moles of H2O:

n = m / MW

n(H2O) = m(H2O) / MW (H2O)

n(H2O) = 91.65 g / 18 g/mol

n(H2O) = 5.092 mol

We calculate the total moles of solution:

total moles of solution = n(H2SO4) + n(H2O)

total moles of solution = 0.1811 mol + 5.092 mol

total moles of solution = 5.2731 mol

With the moles of solution we can calculate the mole fraction of each component:

mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution

mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol

mole fraction (H2SO4)= 0.034

mole fraction (H2O)= moles of (H2O) / total moles of solution

mole fraction (H2O)= 5.092 mol / 5.2731 mol

mole fraction (H2O)= 0.966

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

Solvent
Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

8 0

2 years ago