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3 years ago

Which is easier to remove: a valence electron from, Li or Na? Explain why.

Chemistry

1 answer:

scZoUnD [109]3 years ago

7 0

The atomic number of Li is 3

Electron configuration of Li : 1s² 2s¹

The atomic number of Na is 11

Electron configuration of Na : 1s²2s²2p⁶3s¹

Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.

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fomenos

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3 years ago

What is the pH of a solution with a hydrogen ion concentration of 4.5 x<br> 10-6 M ?

Fittoniya [83]

Answer:

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ {H}^{+} ]$

We have

$pH = - log(4.5 \times {10}^{ - 6} ) \\ = 5.346787...$

We have the final answer as

Hope this helps you

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2 years ago

Water has a density of 1g/ml. what is the mass of the water if it fills a 10ml container?

notka56 [123]

It's 10.
Mass = density x volume
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8 0

2 years ago

Read 2 more answers

How much energy, in joules, does 150.0 g of water with an initial temperature of 25 C need to absorb be raised to a final temper

satela [25.4K]

Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

Initial temperature = 25 °C

Final temperature = 75 °C

Energy absorbed = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 75 °C - 25 °C

ΔT = 50 °C

now we will put the values in formula

q = m . c . ΔT

q = 150 g × 4.186 J/g.°C × 50 °C

q = 31395 J

so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .

5 0

2 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