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4 years ago

Match these items with their examples:

Chemistry

2 answers:

Komok [63]4 years ago

8 0

The correct answers are given below:

1. Salt water is a liquid solution.

A substance is defined as a liquid solution if it is an homogeneous system in which a solute is dissolved in a liquid solvent. For a liquid solution, more than one compound exist in only one phase. Examples of liquid solution are salt dissolved in water and ethanol dissolved in water.

2. Sterling silver is a solid solution.

A solid solution is a solution that contain a minor constituent, which is distributed uniformly within the crystal lattice of the major component. It is a solid state solution of one or more solutes that are dissolved in a solvent. Thus, a mixture of two crystalline solids exist together in a new crystalline solid.

3. Sugar is a compound.

A compound is said to be formed when two or more elements combine together chemically in order to form a new substance. Such a reaction is usually irreversible and a new substance is usually formed. Sugar such as glucose is consider to be a compound because it was formed from the chemical combination of carbon dioxide, water and energy from the sun.

4. Air is a gaseous solution.

A substance is said to be a gaseous solution if two or more gases are mixed together in such a way that they become uniformly mixed up. In this case both the solvent and the solute are gases. Gases can only dissolve in gases if they are mixed under specific conditions. An example of gaseous solution is air, in which oxygen and other gases are dissolved in nitrogen.

5. Gold is an element.

An element refers to a substance that is made from only one type of atom. There are different types of elements in existence and gold is one of them. An element is usually made up of the same number of proton and electron, this makes the element to be neutral. An element will undergo chemical reactions with other element based on its chemical properties.

ehidna [41]4 years ago

8 0

1. salt water
2.sterling silver
3.sugar
4.air
5.gold

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viktelen [127]

(a) f = 5.00 × 10²⁰ Hz, E = 3.32 × 10⁻¹³ J;

(b) f = 1.20 × 10¹⁰ Hz, E = 7.96 × 10⁻²⁴J.

<h3>Explanation</h3>

What's the similarity between a gamma ray and a microwave?

Both gamma rays and microwave rays are electromagnetic radiations. Both travel at the speed of light at $3.00 \times 10^{8}\;\text{m}\cdot\text{s}^{-1}$ in vacuum.

$f = \dfrac{c}{\lambda}$

where

f is the frequency of the electromagnetic radiation,
c is the speed of light, and
$\lambda$ is the wavelength of the radiation.

(a)

Convert all units to standard ones.

$\lambda = 0.600\;\text{pm} = 0.600 \times 10^{-12} \;\text{m}$ .

The unit of $f$ shall also be standard.

$f = \dfrac{c}{\lambda} = \dfrac{3.00\times 10^{8}\;\text{m}\cdot\text{s}^{-1}}{0.600\times 10^{12}\;\text{m}} = 5.00 \times 10^{20}\;\text{s}^{-1}= 5.00\times 10^{20}\;\text{Hz}$ .

For each particle,

$E = h\cdot f$ ,

where

$E$ is the energy of the particle,
$h$ is the planck's constant where $h = 6.63\times 10^{-34}\;\text{J}\cdot\text{s}^{-1}$ , and
$f$ is the frequency of the particle.

$E = h \cdot f = 6.63\times10^{-34}\;\text{J}\cdot\text{s}\times 5.00\times 10^{20}\;\text{s}^{-1} = 3.32\times10^{-13}\;\text{J}$ .

(b)

Try the steps in (a) for this beam of microwave with

$\lambda = 2.50 \;\text{cm} = 2.50\times 10^{-2}\;\text{m}$ .

Expect the following results:

$f = 1.20\times 10^{10}\;\text{Hz}$ , and
$E = 7.96\times 10^{-24}\;\text{J}$ .

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3 years ago

Balance the equation.<br>P + 02 --> P4O10

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Answer:

4P + 5O₂ —> P₄O₁₀

Explanation:

From the question given above, we obtained:

P + O₂ —> P₄O₁₀

The above equation can be balance as illustrated below:

P + O₂ —> P₄O₁₀

There are 4 atoms of P on the right side and 1 atom on the left side. It can be balance by 4 in front of P as shown below:

4P + O₂ —> P₄O₁₀

There are 10 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 5 in front of O₂ as shown below:

4P + 5O₂ —> P₄O₁₀

Now the equation is balanced.

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3 years ago

The periodic table organize elements based on

e-lub [12.9K]

The elements are arranged by their atomic number but are also grouped by what type of element they are.

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4 years ago

which of these pieces of equipment would be the most appropriate for precisely measuring 29 mL of liquid? Explain your reasoning

Anna007 [38]

Answer:

The best equipment would be the graduated cylinder. Why?

Firstly, the smallest marking on the graduated cylinder is 2 mL, while on all the others the smallest marking is way above that, like 25 mL and 100 mL.

Without even going into the details, we can first rule out the volumetric flask, since its smallest marking is 100 mL and even that is already bigger than our sample size, hence we would have no markings to accurately measure out 29 mL of our sample had we used the volumetric flask.

Next to be ruled out would be the Erlenmeyer flask, as you can see in the image, it only has three marking, and as the smallest marking is 25 mL, each marking is at least 25 mL, and even so far as going up to 50 mL. This cannot let us accurately measure 29 mL out at all, due to the markings being way too big to do that. Hence, the Erlenmeyer flask is ruled out.

Finally, the beaker seems to be a worthy candidate! Unfortunately, for the same reason as the Erlenmeyer flask, as you can see in the image each marking represents 10 mL. We cannot measure 9 mL in the beaker accurately, and hence the beaker is ruled too.

We are left with the graduated cylinder, and that is our answer.

Explanation:

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