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Goshia [24]
3 years ago
13

How many atoms of Al(OH)3 in a 72 gram sample? please show work​

Chemistry
1 answer:
Lostsunrise [7]3 years ago
4 0

Answer:

5.559*10^(23) atoms

Explanation:

Molar Mass of Al(OH)3 = 78 g/mol

(72g/78g/mol)*(6.022*10^(23)atoms/mol) = 5.559*10^(23) atoms in 72 grams of Al(OH)3

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Hydrochloric acid, or HCl, reacts with solid NaOH
yuradex [85]

Answer:

NaCI and H20


To simplify it.

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3 years ago
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What is the gravitational potential energy associated whit the spoon at the height relative to the surface of the table.If a spo
igomit [66]

Answer:

0.062J

Explanation:

Given parameters:

Height of raising the spoon = 21cm

Mass of spoon  = 30g

Unknown:

Gravitational potential energy  = ?

Solution:

The gravitational potential energy of a body is the energy by virtue of the position of a such a body in space.

It is given as:

   G. P. E  = mgh

where m is the mass

          g is the acceleration due to gravity

          h is the height

Convert the parameters to appropriate units.

      21cm to m gives 0.21m

      30g to kg gives 0.03kg

   G.P.E  = 0.03 x 9.8 x 0.21  = 0.062J

4 0
3 years ago
What happens when molecules absorb energy?
Nata [24]

Answer:

C. The molecules move slower

Explanation:

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3 years ago
How many moles are in 2.04 x 1024 molecules of H20?
DochEvi [55]

Answer:

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Explanation:

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4 0
2 years ago
Calculate the mass of MgCO3 (84.31 g/mol) precipitated by mixing 10.0 mL of a 0.300 M Na2CO3 solution with 6.00 mL of 0.0400 M M
I am Lyosha [343]

Answer:

m_{MgCO_3}=0.0202molMgCO_3

Explanation:

Hello,

In this case, for this purpose we first have to write the undergoing chemical reaction:

Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3

Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:

n_{Na_2CO_3}=0.010L*0.300\frac{molNa_2CO_3}{1L}=0.003molNa_2CO_3 \\\\n_{Na_2CO_3}^{consumed}=0.006L*0.0400\frac{molMg(NO_3)_2}{1L}*\frac{1molNa_2CO_3}{1molMg(NO_3)_2} =0.00024molNa_2CO_3

In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:

m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3}  \\\\m_{MgCO_3}=0.0202molMgCO_3

Regards.

7 0
3 years ago
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