Ask question

Ask question

All categories

4 years ago

7

Iron is .....

1 answer:

zhannawk [14.2K]4 years ago

8 0

Based on its location on the periodic table, iron is a transition metal. Metals have freely moving electrons that make them great conductors.

Therefore, the answer must be “A transition metal and won’t conduct electricity”

You might be interested in

How do you change Celsious to Faranheight (or however u write them)

Minchanka [31]

Multiply by 1.8 and then add 32.

5 0

3 years ago

Read 2 more answers

Recording station x has determined that the epicenter of a recent earthquake was 250 km away. no information is available yet fr

KatRina [158]

The epicenter was located somewhere on a circle centered at Recording station X, with a radius of 250 km.<span>
</span>

7 0

3 years ago

Read 2 more answers

Write the uses of all elements which are given in the periodic table

Karolina [17]

Answer:

write the uses of all elements which are given in the periodic table

6 0

3 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

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

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

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

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

B) How many moles of Na2O can be produced from 13.5 moles Na?

dolphi86 [110]

Answer: 6.75 moles of $Na_2O$ can be produced from 13.5 moles Na

Explanation:

The balanced chemical equation for formation of $Na_2O$ from Na is:

$4Na+O_2\rightarrow 2Na_2O$

According to stoichiometry :

4 moles of Na give = 2 moles of $Na_2O$

Thus 13.5 moles of $Na$ give = $\frac{2}{4}\times 13.5=6.75moles$ of $Na_2O$

Thus 6.75 moles of $Na_2O$ can be produced from 13.5 moles Na

3 0

3 years ago