Help me with this one please!

How many moles of iron can be made from 3 miles of Fe2O3

denis23 [38]

Answer: two moles

Explanation:

In one mole of fe2o3 there are two moles of iron atoms. Now gram atomic wt. of Fe is 56.So in one mole of Fe2O3 there are 2×56=112 gm of Fe. Hence in 1.75molesFe2O3 there are 1.75×2 moles or 1.75×56 gm of iron

8 0

3 years ago

Plz help me thank you so much for helping

malfutka [58]

The answer to it is A

6 0

3 years ago

Read 2 more answers

What is the pOH of a solution that has a OH - concentration equal to 1.3x 10^-10

Paul [167]

Answer: 9.88 (approx10)

Explanation:

The pOH of a solution is the negative logarithm of the hydroxide-ion concentration

7 0

3 years ago

Read 2 more answers

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

3 years ago

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C 6H 5COOH) and 0.20 M sodium benzoate (C 6H 5COONa),

Zigmanuir [339]

Answer:

pH = 4.05

Explanation:

The pH of the benzoic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.187

pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]

Where [] can be understood as moles of each specie.

Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.

Initial moles:

Initial moles of benzoic acid and sodium benzoate are:

Acid: 250mL = 0.250L ₓ (0.250 moles / L) = 0.0625 moles of benzoic acid

Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = 0.050 moles of sodium benzoate

Moles after reaction:

Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:

HCl + C₆H₅COONa → C₆H₅COOH + NaCl

That means after reaction moles of both species are:

Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles

Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles

Replacing in H-H equation:

pH = 4.187 + log [0.0475] / [0.065]

pH = 4.05

5 0

3 years ago