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Lena [83]
3 years ago
7

How many Al atoms are there in 40.5 g of Al foil? Al:27

Chemistry
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

c) 9.03 x 10^23

Explanation:

find the molar mass of Al

Al is 27.0 grams

Then use that, to find the number of moles in Aluminum.

Then use Avogadro's number which is 6.02 * 10^23

After that, write all of that down with dimensional analysis.

40.5 g * 1 mol/ 27.0 g of Al  * 6.02 x 10^23 / 1mol

As your final answer, you will get 9.03 * 10^23 atoms with sig figs.

Hope it helped!

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Sodium metal and water react to form hydrogen and sodium hydroxide. If 1.99 g of sodium react with water to form 0.087 g of hydr
zavuch27 [327]

Answer:

1.56 g of water was involved in the reaction

Explanation:

From the stoichiometric equation

2Na + 2H2O = 2NaOH + H2

NB : Mm Na= 23, Mm H2O = ( 2+16)= 18

2(23) of Na requires 2(18) of water

Hence 1.99g of Na will require 1.99×2×18/2(23) of water = 1.56 g of water

7 0
4 years ago
Read 2 more answers
Please answer this question ASAP
Nezavi [6.7K]

Answer:

put a test tube over the opening, remove it and quickly put a lit splint near the mout or in the tube. if you hear a squeaky pop it is hydrogen.

Explanation:

hydrogen ignites in air.

4 0
3 years ago
(a) if a sample containing 2.00 ml of nitroglycerin is detonated, how many total moles of gas are produced? (b) if each mole of
KATRIN_1 [288]
<span>(2.09 mL) x (1.592 g/mL) / (227.0871 g C3H5O9N3/mol) = 0.014652 mole C3H5O9N 4 moles C3H5O9N produce 12 + 6 + 1 + 10 = 29 moles of gases, so: (0.014652 mole C3H5O9N) x (29/4) = 0.106 mole of gases (b) (0.106 mol) x (46 L/mol) = 4.88 L gases (c) (0.014652 mole C3H5O9N) x (6/4) x (28.0134 g/mol) = 0.616 g N2</span>
3 0
3 years ago
Which of the following is a galvanic cell?
Gemiola [76]

Answer:

i think Aluminum (Al) oxidized, zinc(Zn) reduced

5 0
3 years ago
Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th
aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = ?

Depression in freezing point = \Delta T_f

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample  = 0.275 mol/kg

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  non electrolyte)

\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg

\Delta T_f=11^oC

\Delta T_f=T- T_f

T_f=T- \Delta T_f=179^oC-11^oC=168^oC

168°C is the melting point of your impure sample.

4 0
3 years ago
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