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jekas [21]
3 years ago
9

An element has a metallic-gray appearance. It also has the properties of both a metal and a nonmetal. In which section of the ta

ble would you find this element?
Select one:

3

4

1

2

Chemistry
2 answers:
Rudiy273 years ago
4 0

Answer:

The answer is where the metalloids are located

Explanation:

katrin2010 [14]3 years ago
3 0

Answer:

you would find it in section 3 as these are semi-metals

Explanation:

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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

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3 years ago
How many moles of a gas would occupy 22.4 Liters at 273 K and 1 atm?<br><br><br> Read bottom comment
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Answer:

The following relationship makes this possible: 1 mole of any gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.

Explanation:

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Increasing substrate concentration also increases the rate of reaction to a certain point. Once all of the enzymes have bound, any substrate increase will have no effect on the rate of reaction, as the available enzymes will be saturated and working at their maximum rate.

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