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NeX [460]
3 years ago
7

What is the molar mass of Al(NO3)3 · 3H2O?

Chemistry
1 answer:
Nana76 [90]3 years ago
3 0

Answer:

Go to your periodic table and look for each element. Find the mass for Aluminum, Nitrogen, Oxygen, and Hydrogen.

After you find those, for each compound you must add them together and find the least amount of sig figs after the decimal point.

For the first compound, Al(NO3)3, you will have a total of

1 Aluminum atom

3 Nitrogen atoms

9 Oxygen atoms

Aluminum has 26.982 g/mol

Nitrogen has 14.007 g/mol

Oxygen has 15.999 g/mol

Now multiply those numbers by the amount of atoms of each element.

(1)26.982 g/mol = 26.982 g/mol

(3)14.007 g/mol = 42.021 g/mol

(9)15.999 g/mol = 143.991 g/mol

now add all of those numbers together, and you see that their least significant figure after the decimal is 3, so round to 3 digits after the decimal point.

26.983 + 42.021 + 143.991 = 212.994

now do the same for the other compound

to start you off.. you have

6 Hydrogen atoms

3 Oxygen atoms

Hydrogen has 1.009 g/mol

Oxygen has 15.999 g/mol

the least significant figure after the decimal point is 3, so round you 3 digits after the decimal point.

after you finish getting your totals, you. until them and find the greatest sig fig over all. comment on this if you need further instruction :)

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For parts a & b below, derive only the initial value problem set up.
Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )

The rate of the sugar coming into the tank is 0

\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min

So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100}  \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C  \\ \\  \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So;  y(t) = 25 e^{-\dfrac{t}{100}

Thus, the amount of sugar at any time t = \mathbf{25 e^{^{-\dfrac{t}{100}}}}

B) For tank B :

\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\  \dfrac{dy}{50-y }= \dfrac{dt}{50}

8 0
3 years ago
How do we name chemicals
Vesnalui [34]
The answer is:

You take the second element and you add "ide".

I hope this helps! 

3 0
3 years ago
BaCl2+Al(NO3)3=Ba(NO2)3+AlCl3
Blizzard [7]

Answer:

6.75 moles of Ba(NO₃)₂.

Explanation:

The balanced equation for the reaction is given below:

3BaCl₂ + 2Al(NO₃)₃ —> 3Ba(NO₃)₂ + 2AlCl₃

From the balanced equation above,

2 moles of Al(NO₃)₃ reacted to produce 3 moles of Ba(NO₃)₂

Finally, we shall determine the number of mole of Ba(NO₃)₂ produced by the reaction of 4.25 moles of Al(NO₃)₃. This can be obtained as illustrated below:

From the balanced equation above,

2 moles of Al(NO₃)₃ reacted to produce 3 moles of Ba(NO₃)₂.

Therefore, 4.25 moles of Al(NO₃)₃ will react to produce = (4.50 × 3)/2 = 6.75 moles of Ba(NO₃)₂.

Thus, 6.75 moles of Ba(NO₃)₂ were obtained from the reaction.

7 0
3 years ago
How many valence electrons does an atom of any halogen have?
Makovka662 [10]
All halogens are stable so they have 8 electrons in their last shell
7 0
3 years ago
What is the mass ratio of aluminum to oxygen in aluminum oxide, Al2O3?
liubo4ka [24]

Answer:

9 : 8

Explanation:

Aluminum oxide has the following formula Al₂O₃.

Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:

Mass of Al in Al₂O₃ = 2 × 27 = 54 g

Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g

Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:

Mass of Al = 54 g

Mass of O₂ = 48 g

Mass of Al : Mass of O₂ = 54 : 48

Mass of Al : Mass of O₂ = 54 / 48

Mass of Al : Mass of O₂ = 9 / 8

Mass of Al : Mass of O₂ = 9 : 8

Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8

5 0
3 years ago
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