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Vikentia [17]
3 years ago
8

Give four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound e. phys

ical change d. element f. chemical change
Chemistry
1 answer:
zzz [600]3 years ago
6 0

1. homogenous: sugar solution

2. heterogeneous: sand solution

3. compound: water

4. physical change: ice melting

5. element: hydrogen

6. chemical change: burning fire

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What are some examples of some monotonic ions found in the body and how do they affect the body? Please help, anything will help
Luba_88 [7]
Atoms are electrically neutral because the number of protons, which carry a 1+ charge, in the nucleus of an atom is equal to the number of electrons, which carry a 1- charge, in the atom. The result is that the total positive charge of the protons cancels out the total negative charge of the electrons so that the net charge of the atom is zero. Most atoms, however, can either gain or lose electrons; when they do so, the number of electrons becomes different from the number of protons in the nucleus. The resulting charged species is called an ion.

4 0
3 years ago
Consider an electron placed in a region of space in which the electric potential spatially. The electron is placed at a point wh
loris [4]

Answer:

1.33\cdot 10^7 m/s

Explanation:

For a charged particle accelerated by an electric field, the kinetic energy gained by the particle is equal to the decrease in electric potential energy of the particle; therefore:

K_f-K_i = -q\Delta V

where

K_f is the final kinetic energy

K_i is the initial kinetic energy

q is the charge of the particle

\Delta V is the potential difference

In this problem,

q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-500 V-(-1000 V)=500 V

The electron starts from rest, so its initial kinetic energy is

K_i=0

Therefore,

K_f=-(-1.6\cdot 10^{-19})(500)=8\cdot 10^{-17}J

We can write the final kinetic energy of the electron as

K_f=\frac{1}{2}mv^2

where

m=9.11\cdot 10^{-31} kg is the electron mass

v is the final speed

And solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(8\cdot 10^{-17})}{9.11\cdot 10^{-31}}}=1.33\cdot 10^7 m/s

5 0
3 years ago
What would be the mass of a 33.5dm3 sample of O2 at STP?
liraira [26]

Answer:

• One mole of oxygen is equivalent to 16 grams.

→ But at STP, 22.4 dm³ are occupied by 1 mole.

{ \tt{22.4 \:  {dm}^{3}   \: \dashrightarrow \: 16 \: grams}} \\  { \tt{33.5 \:  {dm}^{3}  \:  \dashrightarrow \: ( \frac{33.5 \times 16}{22.4} ) \: grams}} \\  \\  \dashrightarrow \: { \boxed{ \tt{23.94 \: g \approx \: 24 \: grams}}}

7 0
2 years ago
Warning signs are sometimes placed on aerosol cans to prevent people from throwing them into a fire. What would be true about th
GarryVolchara [31]

Answer:

The correct option is volume stays constant

Explanation:

When a gas container (in this case an aerosol can) is subjected to heat (from fire), the temperature of the can and subsequently <u><em>the temperature of the gas itself increases</em></u>, an increase in the temperature of the gas cause <u><em>the pressure to also increase;</em></u> as the gas molecules will collide more and faster with each other and against the wall of the can. However, the volume of the gas will remain the same as before it was subjected to the heat - the gas particles do not get destroyed or increased as a result of the heat (law of conservation of matter explains this).

3 0
3 years ago
Read 2 more answers
What is the transfer of electrons in Al + Cl = AlCl3
otez555 [7]

Answer:

3 e⁻ transfer has occurred.

Explanation

This is a redox reaction.

  • Oxidation (loss of electrons or increase in the oxidation state of entity)
  • Reduction (gain of electrons or decrease in the oxidation state of the entity)
  • An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
  • [Ne]= (1s²) (2s² 2p⁶)

A combination of both the reactions( Half-reactions) leads to a redox reaction.

Let us look at initial configurations of Al and Cl

[Al]= 1s² 2s² 2p⁶ 3s² 3p¹

[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵

Hence, Al can lose 3 electrons to achieve octet config.

and, Cl can gain 1e to achieve nearest noble gas config. [Ar]

This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.

Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃

Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)

Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.

3 0
3 years ago
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