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3 years ago

Blackmail threat of informational disclosure is an example of which threat category?

Chemistry

1 answer:

Rashid [163]3 years ago

6 0

Answer:

Information Extortion.

Explanation:

Computer oriented crime also known as cyber crime that intentionally harm the victims. The complete data of the company or net banking information can be hacked easily by the cyber crime.

Different category of the threats are included in the cyber crime. The information extortion threat is the crime in which the hackers hack complete information and control the data of the victim. The criminal an black mail the victim regarding the hacked data.

Thus, the correct answer is information extortion.

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Which of the following is a unit for volume? m3 km cg dkm

babymother [125]

M³ is a unit for volume.

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3 years ago

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If the first isotope has a mass of 259.98 amu and a percent abundance of 35%, what is the percent abundance of the second isotop

brilliants [131]

If it is assumed that there are only two isotopes then the percent abundance needs to add up to 100%

100-35= 65%

The second isotope will have a 65% abundance.

7 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

What is the formula to use for this question, "how far does light travel in 2.5s"?

Nutka1998 [239]

The definition of the speed of light is exactly 299,792,458 meters per second, so to find how far it travels in a time period, multiply the speed of light times the time. Aka c=299,792,458m/s where c is speed of light, m is meters, and s is seconds. So for example to find how far light travels in 5 seconds, multiply by 5.

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3 years ago

Activity: A Ferris wheel with a diameter of 60.0 m is moving at a speed of 2.09 m/s. What is the centripetal

GarryVolchara [31]

Answer:

Centripetal acceleration of the wheel is $0.145\ m/s^2$ .

Explanation:

We have,

Diameter of a Ferris wheel is 60 m

Radius of the wheel is 30 m

Speed of the wheel is 2.09 m/s

It is required to find the centripetal acceleration of the wheel. The formula of centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2.09)^2}{30}\\\\a=0.145\ m/s^2$

So, the centripetal acceleration of the wheel is $0.145\ m/s^2$ .

5 0

3 years ago