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2 years ago

What is the difference between iconic bonds and covelent bonds I NEED HELP NOW!!!!!!!

Chemistry

2 answers:

Hatshy [7]2 years ago

8 0

Well to put it simply iconic bonds give electrons to another atom while covalent bonds share electrons.

jolli1 [7]2 years ago

3 0

An ionic bond is a chemical link between two atoms caused by the electrostatic force between oppositely-charged ions in an ionic compound.

A covalent bond is a chemical link between two atoms in which electrons are shared between them

I hope this helps

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3 years ago

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Gold (Au) is considered a(n) Compound Potatoes Mixture Element

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2 years ago

A sample of 25mls of HCL was mixed with 25mls of KOH of the same concentration in a calorimeter. As a result the temperature ros

xxTIMURxx [149]

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3 years ago

Gallium has two naturally occurring isotopes with the following masses and natural abundances: Isotope Mass ( amu ) Abundance (%

ICE Princess25 [194]

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Ar = ∑(mass * abundance) / 100
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3 years ago

A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or

Pavlova-9 [17]

Answer:

(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.

(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.

Explanation:

Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.

In 300 years we have 600,000 hours.

With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.

If we have 2000 work hours a year, it is expected 0.0033 deaths/year.

$1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year$

The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).

In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:

$FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8} hours=1.67*10^{2} =167$

A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.

If we assume FAR=4, as in an average chemical plant, we expect

$4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}$

This is equivalent to say

$\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}$

The expected number of fatalities on a average chemical plant are one in 12500 years.

4 0

3 years ago