What is au electron configuration

How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?

Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

( V×C )acid = ( V×C )base

∴ C H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ C KOH = 0.0050 N = 0.0050 eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ C KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×C ) acid / C KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0

3 years ago

How are protons, electrons, and neutrons correctly compared

Fiesta28 [93]

they all have one thing in common and that its all made up of atoms. When these components are active it creates energy

4 0

3 years ago

400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution

ch4aika [34]

Answer: The molecular weight of protein is $1.14\times 10^2g/mol$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of protein = 400 mg = 0.4 g (Conversion factor: 1 g = 1000 mg)

$M_{solute}$ = molar mass of protein = ?

$V_{solution}$ = Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol$

Hence, the molecular weight of protein is $1.14\times 10^2g/mol$

4 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

A sample of nitrogen gas is produced in a reaction and collected under water in a graduated cyliner. The temperature is 26.3 oC

Rudiy27

The mass of nitrogen collected is mathematically given as

M-N2=0.025gram

<h3>What is the mass of nitrogen collected?</h3>

Question Parameters:

A sample weighing 2.000g

the liberated NH3 is caught in 50ml pipeful of H2SO4 (1.000ml = 0.01860g Na2O).

T=26.3c=299.3K

Pressure=745mmHg=745torr

Pressure of N2=745-25.2=719.8torr

Generally, the equation for the ideal gas is mathematically given as

PV=nRT

Therefore

719.8/760=45.6/1000=n*0.0821*299.3

n=0.00176*14

In conclusion, the Mass of N2

M-N2=0.00176*14

M-N2=0.025gram