All categories

3 years ago

Calculate the average mass for iron. 15.2% iron-55, 85.8% iron-56

Chemistry

1 answer:

Roman55 [17]3 years ago

5 0

Answer:

Iron 55 (Fe-55) is the iron isotope whose nucleus consists of 26 protons and 29 neutrons. It is a radioisotope that disintegrates by electron capture in manganese 55.

Explanation:

i hope this helps if not tell me

You might be interested in

A 2.5 mol sample of phosphorus pentachloride, PCl5 dissociates at 160C and 1.00atm to give 0.338 mol of phosphorus trichloride a

Readme [11.4K]

Explanation:

Moles of phosphorus pentachloride present initially = 2.5 mol

Moles of phosphorus trichloride at equilibrium = 0.338 mol

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Initially

2.5 mol 0 0

At equilibrium:

(2.5 - x) mol x x

So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol

Mass of 0.338 moles of phosphorus trichloride at equilibrium:

= 0.338 mol × 137.5 g/mol = 46.475 g

Moles of phosphorus pentachloride present at equilibrium :

= (2.5 - 0.338) mol = 2.162 mol

Mass of 2.162 moles of phosphorus pentachloride at equilibrium:

= 2.162 mol × 208.5 g/mol = 450.777 g

Moles of chloride gas present at equilibrium : 0.338 mol

Mass of 0.338 moles of chloride gas at equilibrium:

= 0.338 mol × 71 g/mol = 23.998 g

3 0

3 years ago

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

<u>Answer:</u> The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago

What is the name of the molecule shown below?

DiKsa [7]

Answer:

the correct awnser is b 1-heptene

8 0

3 years ago

What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters?

Leto [7]

7.54 atm kf im correct

3 0

3 years ago

In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in

RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

Density of water = 1000 $kg/m^{3}$

Therefore, mass of water = Density × Volume

= $1000 kg/m^{3} \times 0.25 m^{3}$

= 250 kg

Initial Temperature of water ( $T_{1}$ ) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ( $dH_{v}$ ) at $140^{o}C$ is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ( $140^{o}C$ ) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

= 62.5 (kg) × 2133 (kJ/kg)

= $133.3 \times 10^{3}$ kJ

All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

= 125520 kJ

Time required = Heat required / Power rating

= $\frac{125520}{37}$

= 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

$\frac{3392 sec}{60 sec/min}$

= 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0

3 years ago