Convert 0.60 moles of Fe to moles of Fe2O3

Explanation:

The answer would be B.

As paramagnetic with 3 unpaired electrons. Since there are 6 ligands around the Co+2 ion it isoctahedral and these ligands are neutral. This makes the overall charge on the complex +2 and therefore comes from the configuration for Co+2 which is [Ar] 3d7. Since it is in high spin you must fill all the orbitals with at least one electron and then pair up any that remain. If you do this, 3 unpaired electrons remain. Para magnetism occurs in substances with unpaired electrons.

3 0

3 years ago

A wooden artifact from a Chinese temple has a 14C activity of 42.8 counts per minute as compared with an activity of 58.2 counts

Arlecino [84]

Answer : The age of the artifact is, $2.54\times 10^3\text{ years}$

Explanation :

Half-life = 5715 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{5715\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time taken to decay.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken by sample = ?

a = initial activity of the reactant = 58.2 counts per minute

a - x = activity left after decay process = 42.8 counts per minute

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{58.2}{42.8}$

$t=2540.5\text{ years}=2.54\times 10^3\text{ years}$

Therefore, the age of the artifact is, $2.54\times 10^3\text{ years}$

4 0

3 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

3 years ago

When the fuel mixture contained in a 1.49 l tank, stored at 750 mmhg and 298 k, undergoes complete combustion, how much heat is

Ket [755]

Get to know first how many moles in the gas:n = pV/RT= (1.013*10^5*750/760) Pa *1.49*10^-3 m^3/(8.314 J/(molK)*298) n = 0.0601 moles.
The combustion energies are 889 kJ/mol (methane) and 2 220 kJ (propane) x = moles methane, y = moles propane
x*889 + y*2220 = 778 x + y = 0.0601----------- x = 0.267784 moles = 0.267784*100/0.0601 = 44.6 % y = 0.243216 moles = 0.243216*100/0.0601 = 55.4 %

4 0

2 years ago

O

baherus [9]

The answer is the first one I think

8 0

2 years ago