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gulaghasi [49]
4 years ago
9

Someone please help with this

Chemistry
1 answer:
ASHA 777 [7]4 years ago
7 0

Answer: the mass of the carbem 601 p1

Explanation:

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If we place 15 moles of NaN3, how many<br> grams of NaN3 do you have before the<br> reaction?
Anon25 [30]

Answer:

25.8 g C6H12O6

Explanation:

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3 years ago
Describe the observation Robert Brown made.​
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In 1827, Brown observed, using a microscope, that small particles ejected from pollen grains suspended in water executed a kind of continuous and jittery movement, this was named “Brownian motion”. ... This random movement of particles suspended in a fluid is now called after him.

Explanation:

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3 years ago
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mina [271]
If you would draw the Lewis structures of these atoms, you would see that A has 2 electron pairs and 2 lone electrons (that can bond). For B you’d see that you only have 1 electron that can form a bond. This means that 1 atom of A (2 lone electrons) can bond with 2 atoms of B. To know the kind of bond you have to know wether or not there will be a ‘donation’ of an electron from one atom to another. This happens when the number of electrons on one atoms is equal to the number of electrons another atom needs to reach the noble gas structure. As you can see, this is not the case here. This means that you get an AB2 structure with covalent character.
6 0
3 years ago
The mathematical relationship between gas solubility and pressure is called Henry's Law, solubility = kHPgas where kH is the Hen
Diano4ka-milaya [45]

Answer : The unit of k_H in mol/L.mm Hg is, 1.8\times 10^{-6}mol/L.mmHg

Explanation :

As we know that the k_H is the Henry's Law constant for argon at 25^oC is, 1.4\times 10^{-3}mol/L.atm

Now we have to determine the unit of k_H in mol/L.mm Hg

Conversion used for pressure from atm to mmHg is:

1 atm = 760 mmHg

So,

k_H=1.4\times 10^{-3}mol/L.atm\times \frac{1atm}{760mmHg}

k_H=1.8\times 10^{-6}mol/L.mmHg

Thus, the unit of k_H in mol/L.mm Hg is, 1.8\times 10^{-6}mol/L.mmHg

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3 years ago
The insoluble yellow solid is PbI2, which is?
andreev551 [17]
It is lead (ll) iodide or just lead iodide
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3 years ago
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