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vesna_86 [32]
3 years ago
5

The Law of Conservation of Matter states that in a chemical reaction matter cannot be ___ or ____

Chemistry
1 answer:
valkas [14]3 years ago
8 0

Answer:

The law of conservation of matter states that in a chemical reaction matter cannot be created or destroyed

Explanation:

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A student has a piece of aluminum metal. Which is the most reasonable assumption the student could make about the metal?
igor_vitrenko [27]

Aluminium belongs to 13th group of periodic table. It undergoes oxidation to given Al^+3 .

It is observed that when aluminium is added to a solution of copper sulphate the colour of the solution changes from blue to grey. It is due to formation of grey coloured solution of aluminium sulphate as

2Al^+3  + 3SO4^-2  ---> Al2(SO4)3


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3 years ago
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How many moles of Cu(OH)2 will be produced if given 1 mole<br> KOH.
stiks02 [169]

Answer:

0.032 mole

Explanation:

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7 0
2 years ago
Help!! I have no idea what to do!
Anvisha [2.4K]
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5 0
3 years ago
Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N
Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of NaN_3

\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}

Molar mass of NaN_3 = 65.01 g/mole

\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 2 mole of NaN_3 react to give 3 mole of N_2

So, 0.311 moles of NaN_3 react to give \frac{0.311}{2}\times 3=0.466 moles of N_2

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = ?

n = number of moles N_2 = 0.466 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 22^oC=273+22=295K

Putting values in above equation, we get:

1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K

V=11.3L

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0
3 years ago
Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous gadolinium
Inessa [10]

Answer:

Gd(g) → Gd^{+}(g) + e^{-} , Gd^{+}(g)→ Gd^{2+}(g) + e^{-}, Gd^{2+}(g) → Gd^{3+}(g) + e^{-}

Explanation:

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionization energy is the energy required to remove the valence electron(outermost) from a neutral atom:

Gd(g)→ Gd^{+}(g) + e^{-}

The second ionization energy is the energy required to remove next/second electron from x^{+} ion. The second ionization energy is always higher than the first:

Gd^{+}(g) → Gd^{2+}(g) + e^{-}

The third ionization energy is the energy required to remove third electron from x^{2+} ion:

Gd^{2+}(g) → Gd^{3+}(g) + e^{-}  

5 0
3 years ago
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