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Nady [450]
2 years ago
5

What is a combination of two or more substances in which each substance retains its properties?

Chemistry
1 answer:
garri49 [273]2 years ago
8 0

Answer:

Mixture

Explanation:

Mixture -  

It refers to the composition of various substance in random proportion , where the mixture formed may or may not have uniform composition , and the various components does not lose their identity and there is no chemical change observed , is referred to as mixture.

For example ,

The mixture of sand and chalk powder , where both the substances mix together to form a mixture , but still retain their property.

Hence , from the given statement of the question,

The correct term is mixture.

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Both E. coli and Salmonella are single-celled organisms. They do not have a nucleus or other membrane-bound organelles. Based on
AfilCa [17]

Answer:

domain bacteria

Explanation:

Salmonella and E. coli are same in the sense that they are both bacteria,

7 0
3 years ago
There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc
zhuklara [117]

Answer:

K_{sp} of Zn(CH_{3}COO)_{2} is 0.0513

Explanation:

Solubility equilibrium of Zn(CH_{3}COO)_{2}:

Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}

Solubility product of Zn(CH_{3}COO)_{2} (K_{sp}) is written as-            K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}

Where [Zn^{2+}] and [CH_{3}COO^{-}] represents equilibrium concentration (in molarity) of Zn^{2+} and CH_{3}COO^{-} respectively.

Molar mass of Zn(CH_{3}COO)_{2} = 183.48 g/mol

So, solubility of Zn(CH_{3}COO)_{2} = \frac{43.0}{183.48}M = 0.234M

1 mol of Zn(CH_{3}COO)_{2} gives 1 mol of Zn^{2+} and 2 moles of CH_{3}COO^{-} upon dissociation.

so,   [Zn^{2+}] = 0.234 M and [CH_{3}COO^{-}] = (2\times 0.234)M=0.468M

so, K_{sp}=(0.234)\times (0.468)^{2}=0.0513          

8 0
3 years ago
Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.
Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

\%C_E = \frac{E}{T} * 100\%

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In (Ca(OH)_2)

The amount of all elements is: (i.e formula mass of (Ca(OH)_2))

T = 1 * Ca + 2 * H + 2 * O

T = 1 * 40 + 2 * 1 + 2 * 16

T = 74

The amount of calcium is: (i.e formula mass of calcium)

E = 1 * Ca

E = 1 * 40

E = 40

So, the percentage component of calcium is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{40}{74} * 100\%

\%C_E = \frac{4000}{74}\%

\%C_E = 54.05\%

The amount of hydrogen is:

E = 2 * H

E = 2 * 1

E = 2

So, the percentage component of hydrogen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{2}{74} * 100\%

\%C_E = \frac{200}{74}\%

\%C_E = 2.70\%

Similarly, for oxygen:

The amount of oxygen is:

E = 2 * O

E = 2 * 16

E = 32

So, the percentage component of oxygen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{32}{74} * 100\%

\%C_E = \frac{3200}{74}\%

\%C_E = 43.24\%

5 0
2 years ago
Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun
Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂=\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}=0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

\frac{0.92678}{0.0309}=Si³⁰

Si³⁰=29.99288

8 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 20.4 g g of carbon were burned in the presence
Aleonysh [2.5K]

Answer:

C + O2 -> CO2

C 20.4/12 = 1.7

O 70.4-16.0/16=3.4

CO2 1.7 mol * 44 = 74.8

3 0
3 years ago
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