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3 years ago

Write the IUPAC name OF THE FOLLOWING COMPOUNDS

Chemistry

1 answer:

goldenfox [79]3 years ago

7 0

Answer:

Explanation:

a) 2 chloro butane

b) 2-3 dimethyl butane

c) 2 bromo 3 nitro pentane

d) 2-3 trimethyl pentane

e) 2-bromo,3-methyl,4-nitro hexane

f) 2-methyl cyclo butane

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The atoms and molecules that move the most quickly.

Colt1911 [192]

Plasma
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3 years ago

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The number 0.000402 expressed in scientific notation is

goldfiish [28.3K]

Answer:

hope its help you

Explanation:

4.02 x 10-4

3 0

3 years ago

A wave has a frequency of 3,5MHz and a wavelength of 85,5 m. What is the wave velocity through

Yuki888 [10]

Answer:

velocity = 29925×10⁶ m/s

Explanation:

Given data:

Frequency = 35 MHz

Wavelength = 855 m

Velocity = ?

Solution:

MHz to Hz

35×10⁶ Hz

Formula:

velocity = Wavelength × Frequency

velocity = 855 m × 35×10⁶ Hz

velocity = 29925×10⁶ m/s

Hz = s⁻¹

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3 years ago

How many half-lives would something have gone through if you had 75% daughter product and 25% parent?

levacccp [35]

Answer:

Two Half-lives

Explanation:

Let number of Parent nuclei Initially present be X,

Then, finally $\frac{X}{4}$ Parent nuclei Will remain with $\frac{3X}{4}$ daughter nuclei.

In one half- life , parent nuclei becomes half of initial.

So, starting with X parent nuclei,

After one half-life, it will degrade to $\frac{X}{2}$ .

After another half life , Parent nuclei will become half of $\frac{X}{2}$

Which is equal to $\frac{X}{4}$ .

So, Parent nuclei have to go through Two half-lives.

6 0

3 years ago

Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0

nexus9112 [7]

Answer: The correct answer is Option b.

Explanation:

To calculate the amount of heat absorbed or released, we use the following equation:

$q=mc\Delta T$ .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of water = 4.186 J/g ° C

$\Delta T$ = Change in temperature

For Trial 1:

We are given:

$m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J$

Putting values in equation 1, we get:

$q=30g\times 4.186J/g^oC\times 40^oC$

q = 5023.2 J

For trial 2:

We are given:

$m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J$

Putting values in equation 1, we get:

$q=40g\times 4.186J/g^oC\times 10^oC$

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = $(5023.2-1674.4)J=3347J$

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

4 0

3 years ago