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8_murik_8 [283]
2 years ago
9

What is the difference between chemistry and science

Chemistry
1 answer:
sergeinik [125]2 years ago
8 0

Answer:

hi

Explanation:

chemistry the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.

Science the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

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In a mixture of carbon dioxide in water ( a soft drink ) the Carbon Dioxide is the solute
Nataly [62]

Answer:

true :)

Explanation:

hope this helps :)

4 0
3 years ago
Determine the oxidation number for nitrogenin
iris [78.8K]

Answer:

1(a) N = 3

(b) N = 0

(c) N = 5

(d) N = -2

(2) Molecular formula for benzene is C6H6

Explanation:

1(a) N02 1-

N + (2×-2) = -1

N-4 = -1

N = -1+4 = 3

(b) N2

2(N) = 0

N = 0/2 = 0

(c) NO2Cl

N + ( 2×-2) + (-1) = 0

N - 4 - 1 = 0

N - 5 = 0

N = 0+5 = 5

(d) N2H4

2(N) + (4×1) = 0

2N + 4 = 0

2N = 0 - 4 = -4

N = -4/2 = -2

(2) Molcular mass of benzene = 78g/mole = (6×12g of carbon) + (6×1g of hydrogen) = 72+6 = 78g/mole

Therefore, molecular formula for benzene is C6H6

7 0
3 years ago
Which formula is an empirical formula? (1) CH4 (2) C2 H6 (3) C3 H6 (4) C4 H10
rosijanka [135]
CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).
3 0
3 years ago
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
3 years ago
Gerry is looking at salt under a powerful microscope and notices a crystalline structure. What can be known about the salt sampl
Phoenix [80]
Mined salt has sharp edges v s sea salt with rounded edged
4 0
2 years ago
Read 2 more answers
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