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timofeeve [1]
3 years ago
7

1-bromo-2-chloro-4-propylpentene

Chemistry
1 answer:
Nonamiya [84]3 years ago
3 0
..:)../:................

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An object has a density of 0.73 g/cm3. Will the object sink or float in water?
Galina-37 [17]

Answer:

<h2>It'll float</h2><h2>Less than 1g/cm3 it'll float in water </h2>

4 0
3 years ago
Infrared light from the Sun is detected on Earth as _____.
umka21 [38]

Answer: heat

Explanation:

Infrared radiation is popularly known as "heat radiation",[21] but light and electromagnetic waves of any frequency will heat surfaces that absorb them. Infrared light from the Sun accounts for 49%[22] of the heating of Earth, with the rest being caused by visible light that is absorbed then re-radiated at longer wavelengths

4 0
4 years ago
How many different principal quantum numbers can be found in the ground-state electron configuration of nickel?A) 2.B) 3.C) 4.D)
Naddik [55]

Answer:

C) 4.

Explanation:

Hello!

In this case, since the electron configuration of nickel at its ground-state, considering 28 as its atomic number and the number of electrons it has in one atom, is:

Ni^{28}:1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^8

We can see it has four energy levels, 1, 2, 3 and 4, which are related to the following principal quantum number, that describes the energy of an electron in the atom and its most probable distance with respect to the nucleus.

Therefore, nickel has C) 4 different principal quantum numbers.

Best regards!

5 0
3 years ago
A 64.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 32.0 ml of koh at 25 ∘c.
aleksley [76]
Hello!

The reaction between HBr and KOH is the following:

HBr+KOH→H₂O + KBr

To calculate the amount of HBr left after addition of KOH, you'll use the following equations:

HBr_f=HBr_i-KOH=([HBr]*vHBr)-([KOH]*vKOH) \\  \\ HBr_f=(0,25M*0,64L)-(0,5M*0,32L)=0 mol HBr

That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be neutral, close to 7. 

Have a nice day!
3 0
3 years ago
If a chemist analyzes a 3.84g sample containing sand and table sugar, and recovers 1.43g of      sand, what  percent by mass of
n200080 [17]
We are given with a mixture of sand and table sugar. The mixture has a mas s of 3.84 grams. After recovery, it was found out that there is 1.43 grams of sand. The amount of table sugar then is 3.84-1.43 or 2.41 grams. This is equal to 2.41/3.84 or 62.76% by mass/
4 0
3 years ago
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