What is the total amount of heat in a substance

If an electron loses energy what may happen to it

MAVERICK [17]

When an electron is hit by a photon of lights,it absorbs the quanta of energy the photon was carrying and moves to a higher energy state.Electrons therefore have to jump around within the atom as they either gain or lose energy.

7 0

2 years ago

A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘

hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution, $T_f$ = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

$\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC$

$\Delta T_f=i\times K_f\times m$

$Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}$

$5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})$

On solving we get:

$\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol$ ....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of $Ca(NO_3)_2$ in the mixture:

$\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%$

The mass percentage of calcium nitrate is 31.23%.

5 0

3 years ago

To understand the nature of electric fields and how to draw field lines. Electric field lines are a tool used to visualize elect

Brrunno [24]

Explanation:

The electric field is defined as the change in the properties of space caused by the existence of a positively (+) or negatively (-) charged particle. The electric field can be represented by infinitely many lines from a particle, and those lines never intersect each other. Depending on the type of charge we can see different cases:

Let's say we have a positive charge alone (image 1). The field lines are drawn from the centre of the particle outwards to infinity (in other words, they disappear from the edge of the picture). Meaning the direction of the electric field points outwards the particle.

For a negative charge alone (image 2), the lines come from infinity to the centre, and point towards the particle (i.e. lines appear from the edge of the picture).

Let's see what happens if we have two charges together:

Two positive charges (image 3): Since the charges are of the same type (positive), the particles repel each other. Then the field lines will avoid each other so they do not join. The charge is positive, so lines point outwards.
Two negative charges (image 4): Again, the charges are both negative, so they repel. But they are negative, so the field points inwards.
Negative and positive charges (image 5): They are different charges, so the force between them is attractive. This causes the field lines from both to join. They go out of the positive and come into the negative particle.

Image 6:

The lines are passing through infinite points of the space. If we choose a certain point and measure the electric field, we can see to which direction the electric field points. This is the direction of the electric field vector. It does not matter which point we choose; the electric field vector touches the field line only at this point, which means it is tangent to the field line.

7 0

3 years ago

Suppose the Sun appeared to you 900 times dimmer than it does now. How far away from the Sun would you be? (A) 1/9 AU (B) 3 AU

Harrizon [31]

Answer:

The correct answer is option 'c': 30 AUs

Explanation:

For a spherical wave front emitted by sun with total energy 'E' the energy density over the surface when it is at a distance 'r' from the sun is given by

$e=\frac{E}{4\pi r^{2}}$

This energy per unit area is sensed by observer as intensity of the sun.

Let the initial intensity of sun at a distance $r_{1}$ be $e_{1}$

Thus if the sun becomes 900 times dimmer we have

$e'=\frac{e_{1}}{900}\\\\\frac{E}{4\pi r_{2}^{2}}=\frac{1}{900}\times \frac{E}{4\pi r_{1}^{2}}\\\\\Rightarrow r_{2}^{2}={r_{1}^{2}}\times 900\\\\\therefore r_{2}={r_{1}}\times {30}$

Thus the distance increases 30 times.

5 0

3 years ago

Water that contains a larger than normal amount of __________ is still safe to drink.

Lynna [10]

Answer:

Water that contains a larger than normal amount of magnesium is still safe to drink.

Explanation:

Sulfur, lead and arsenic are harmful metals that are toxic to human health. Although magnesium present at high concentrations (above approximately 250 mg/l each) can have a laxative effect. However, data suggest that consumers adapt to these levels as exposures continue.

4 0

3 years ago

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