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vesna_86 [32]
3 years ago
10

Draw the structures of organic compounds a and b. omit all byproducts

Chemistry
1 answer:
Georgia [21]3 years ago
8 0

81. There is 1 carbon, 2 chlorine and fluorine atoms in Freon 12. To draw them it forms a cross with C in the middle and Cl and F both on the opposite side. 
     


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Write a balanced equation for rusting. (Assume rust becomes Fe+2)
Stells [14]

Answer:

4Fe + 3O2 + 6H2O → 4Fe(OH)3

Explanation:

The chemical formula for rust is Fe2O3 and is commonly known as ferric oxide or iron oxide. The final product is a series of chemical reactions simplified below as- The rusting of the iron formula is simply 4Fe + 3O2 + 6H2O → 4Fe(OH)3. The rusting process requires both the elements of oxygen and water.

8 0
2 years ago
3 organisms that help prevent infections
love history [14]

Infection control is the discipline concerned with preventing nosocomial or healthcare-associated infection, a practical (rather than academic) sub-discipline of epidemiology. It is an essential, though often underrecognized and undersupported, part of the infrastructure of health care. Infection control and hospital epidemiology are akin to public health practice, practiced within the confines of a particular health-care delivery system rather than directed at society as a whole. Anti-infective agents include antibiotics, antibacterials, antifungals, antivirals and antiprotozoals.[1]

Infection control addresses factors related to the spread of infections within the healthcare setting (whether patient-to-patient, from patients to staff and from staff to patients, or among-staff), including prevention (via hand hygiene/hand washing, cleaning/disinfection/sterilization, vaccination, surveillance), monitoring/investigation of demonstrated or suspected spread of infection within a particular health-care setting (surveillance and outbreak investigation), and management (interruption of outbreaks). It is on this basis that the common title being adopted within health care is "infection prevention and control." (got from google

8 0
3 years ago
For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.
Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

7 0
3 years ago
Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu
IrinaK [193]

Answer : The atomic radius for Ti is, 1.45\times 10^{-8}cm

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number (N_{A})=6.022\times 10^{23} mol^{-1}

First we have to calculate the volume of HCP crystal structure.

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times V} .............(1)

where,

\rho = density  = 4.51g/cm^3

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass  = 47.87 g/mole

(N_{A}) = Avogadro's number  

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}

V=1.06\times 10^{-22}cm^3

Now we have to calculate the atomic radius for Ti.

Formula used :

V=6R^2c\sqrt{3}

Given:

c/a ratio = 1.669 that means,  c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

V=6R^2\times (1.669a)\sqrt{3}

V=6R^2\times (1.669\times 2R)\sqrt{3}

V=(1.669)\times (12\sqrt{3})R^3

Now put all the given values in this formula, we get:

1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3

R=1.45\times 10^{-8}cm

Therefore, the atomic radius for Ti is, 1.45\times 10^{-8}cm

3 0
3 years ago
What is the mass of 6.3 moles of Sulfur
bixtya [17]
The molar mass of sulfur is 32 g/mol.
The mass of 6.3 moles of Sulfur is 32×6.3=201.6 grams

<span>♡♡Hope I helped!!! :)♡♡
</span>
6 0
3 years ago
Read 2 more answers
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