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nignag [31]
3 years ago
11

How many atoms of zirconium are in 0.3521 mol of zirconium?

Chemistry
1 answer:
lora16 [44]3 years ago
3 0

Answer:

2.12×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.

Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:

1 mole of zirconium also 6.02×10²³ atoms.

Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.

Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

Now put all the given values in the above formula, we get:

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0
3 years ago
Can we all agree chemistry is hard lol
Dmitry_Shevchenko [17]

Yes, I agree.

Chemistry can be difficult.

5 0
3 years ago
Multicellular organisms are organized into tissues, organs, and organ systems.TRUE OR FALSE​
saul85 [17]

Answer:

True i think..............

5 0
3 years ago
Read 2 more answers
Balance this equation K+b2o3=k2o+b
AVprozaik [17]

Answer:

Explanation:

2K and 2b

5 0
2 years ago
Use IUPAC rules to name this compound. A) nitrogen iodide B) nitrogen triiodide C) mononitrogen iodide D) mononitrogen triiodide
Mariulka [41]

Answer:

Name of the compound is Nitrogen triiodide.

Explanation:

According to the IUPAC rules, to naming of the compound the following formula can be applied.

Prefix + Name of first element + Base name element of second element + Suffix.

The given compound - NI_{3}

Name of first element- Nitrogen

Base name element of second element - Iodine

Suffix = 3 = tri

Here, iodine is in ionic form therefore, it becomes iodide. and then suffix will be added in front of the halogen.

Therefore, name of the compound will be  Nitrogen triiodide..

3 0
3 years ago
Read 2 more answers
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