Calculate the percentage by mass of nitrogen in ptcl2(nh3)2.

Plants, plant life cycles, photosynthesis, plant parts, plant growth is what branch of science

WARRIOR [948]

The study of plants, plant life cycles, photosynthesis, plant parts, etc. is under botany (study of plants).

4 0

3 years ago

How much energy is produced in the creation of 5 grams of "O by the process: (10 pts.) 14N + α. 'H + 170 ("N-14.00307 gmole, α-4

LenKa [72]

Answer : The energy produced is $3.410\times 10^{10}J$

Explanation :

First we have to calculate the moles of $^{17}O$ .

$\text{Moles of }^{17}O=\frac{\text{Mass of }^{17}O}{\text{Molar mass of }^{17}O}=\frac{5g}{16.99913g/mole}=0.294133moles$

Now we have to calculate the mass defect.

The balanced reaction is,

$^{14}N+\alpha \rightarrow ^1H+^{17}O$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=[(n_{^1H}\times M_{^1H})+(n_{^{17}O}\times M_{^{17}O})]-[(n_{^{14}N}\times M_{^{14}N})+(n_{\alpha}\times M_{\alpha})]$

where,

n = number of moles = 0.294133 moles

M = molar mass

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

$\Delta m=[(n_{^1H}\times M_{^1H})+(n_{^{17}O}\times M_{^{17}O})]-[(n_{^{14}N}\times M_{^{14}N})+(n_{\alpha}\times M_{\alpha})]$

$\Delta m=[(0.294133mole\times 1.00783g/mole)+(0.294133mole\times 16.99913g/mole)]-[(0.294133mole\times 14.00307g/mole)+(0.294133mole\times 4.0026g/mole)]$

$\Delta m=0.00037943157g=3.7943157\times 10^{-7}kg$

Now we have to calculate the energy produced.

$Energy=\Delta m\times (c)^2$

$Energy=(3.7943157\times 10^{-7}kg)\times (299792458m/s)^2$

$Energy=3.410\times 10^{10}J$

Therefore, the energy produced is $3.410\times 10^{10}J$

3 0

4 years ago

Pls pls help me me pls

galben [10]

Answer:

Danger

Explanation:

7 0

3 years ago

A sample of gas has a volume of 20.0 mL at STP. What will the volume be if the temperature is changed to 546 K and the pressure

Ostrovityanka [42]

The volume did not change, it remained at 20 ml

<h3>Further explanation</h3>

Given

20 ml a sample gas at STP(273 K, 1 atm)

T₂=546 K

P₂=2 atm

Required

The volume

Solution

Combined gas Law :

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

Input the value :

$\tt \dfrac{1\times 20}{273}=\dfrac{2\times V_2}{546}\\\\V_2=\dfrac{1\times 20\times 546}{273\times 2}\\\\V_2=20~ml$

The volume does not change because the pressure and temperature are increased by the same ratio as the initial conditions (to 2x)

5 0

3 years ago

Which one of the following statements is true?A. Some real gases have lower pressures than that calculated for the ideal gas bec

NNADVOKAT [17]

Answer:

A. Some real gases have lower pressures than that calculated for the ideal gas because of attraction between molecules.

Explanation:

At room temperature and moderately low pressures ideal gases among other assumptions:

I. Do not attract or repel each other.

II. Have negligible volume relative to the volume of the containing vessel.

Thus the ideal gas equation:

PV = nRT

Where P = gas pressure = gas molecules' collision with the walls of the containing vessel.

V = volume occupied by the gas sample.

T = Kelvin temperature of the gas

n = number of mole of the gas present.

R = proportionality constant called the molar gas constant = 0.0821 L. atm/K.mol.

However, at low temperatures and high pressures, gases exhibit real behaviors and the

I. attractive forces between the gas molecules may not be negligible.

II. volume of the gas molecules may not be negligible relative to the volume of the container.

At low temperatures, the average kinetic energy of the gas molecules decreases which prevents the molecules breaking free from their molecular attraction.

At high pressures, the density of the gas increases and the molecules come closer to one another. This increases the intermolecular forces between the gas molecules, increases the number of molecules found per unit volume and lowers their speed moving towards the containing wall thus lowering the pressure the gas would exert than if it were in an ideal behavior.

van der Waals corrected the actual pressure and volume of real gases as follows:

( P + an^2/V^2 ) ( V – nb ) = nRT

Where n^2/V^2 = number of molecules per unit volume.

P = observed pressure.

( V – nb ) = effective volume of the gas.

6 0

4 years ago