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3 years ago

Sodium is a metal, like copper, suggest reasons why sodium cannot be used in electrical wires

1 answer:

7 0

it is highly reactive. this is because it is in the first chemical family it only has one valence electron making it very unstable

hope it helps :P

You might be interested in

In the following reaction, what is the quantity of heat (in kJ) released when 5.87 moles of CH₄ are burned?

IRISSAK [1]

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

The enthalpy of a chemical reaction as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

In this case, the balanced reaction is:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g)

and the enthalpy reaction ∆H° has a value of -802 kJ/mol.

This equation indicates that when 1 mole of CH₄ reacts with 2 moles of O2, 802 kJ of heat is released.

When 5.87 moles of CH₄ are burned, then you can apply the following rule of three: if 1 mole of CH₄ releases 802 kJ of heat, 5.87 moles of CH₄ releases how much heat?

$heat=\frac{5.87 molesof CH_{4}x802 kJ}{1 mol of CH_{4} }$

Finally, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

Learn more:

5 0

3 years ago

Examine the given reaction. NH4NO3(s) → NH4+(aq) + NO3–(aq) ΔH° = 25.45 kJ/mol ΔS° = 108.7 J/mol·K Which of the given is correct

kobusy [5.1K]

Answer:

B)−6,942 J /mol

Explanation:

At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:

ΔG = ΔH - TΔS

Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.

Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):

ΔG = ΔH - TΔS

ΔG = 25450J/mol - 298K×108.7J/molK

ΔG = -6942.6J/mol

Right solution is:

8 0

3 years ago

Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantites.

Bingel [31]

Answer:

A. Kilo , K, multiplication by 1000

B. Centi, c

C. Deci, d

D. Mili, m

E. Mega, M

F. Micro, u

7 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

4 years ago

How have ideas about the atom changed in the last two centureies?

dsp73

In the past a scientist named dalton produced an atomic theory. There were certain problems regarding his views. So, later on scientists like chadwick, rutherford and thompson added some fresh light for the real identification of the atom.
dalton said the atom was the smallest unit and it CANNOT BE DIVIDED ANY FURTHER.
NOTE... this was renected with the discovery of the proton, neutron and electron as the sub atomic particles.

8 0

3 years ago