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2 years ago

Which type of matter is composed or two or more elements that are chemically combined in a fixed proportion?

1 answer:

6 0

Hello.

When dealing with two or more elements in a fixed proportion, your answer is going to be "B.) Compound." The explanation is pretty basic when you narrow it down - think of it like the English version of a compound sentence; there are two independent clauses that have related ideas (such as this one, for example).

Let's use an example of a compound element, such as $H_{2}O$ , or Water.
This is a compound element that is composed of the two elements "Hydrogen" and "Oxygen" - for every 2 Hydrogen atoms there is one Oxygen atom.

I hope this helps!

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In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

How is it possible for helium to have 8 spectral lines when it only has 2 electrons?

neonofarm [45]

It has more lines in it compared to hydrogen emission spectrum. It is mainly because the helium atom has more electrons than a hydrogen atom. Therefore, more electrons get excited when we pass a white light beam through a helium sample, and it causes the emission of more spectral lines

5 0

2 years ago

How do the polar easterlies and the jet stream work together to affect the

ahrayia [7]

Answer:

they can carry cold arctic air, producing snow and freezing weather.

Explanation:

-w- hope it helps

8 0

2 years ago

What is the condition of the outside air at a certain time and place?

11111nata11111 [884]

Weather... weather is the obvious answer

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3 years ago

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what is known as electronegativity and mention how the electro negativity vary long the periods and down the groups..please i ne

Kobotan [32]

Answer:

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.

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2 years ago

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