Definition: This is a measure of the quantity of matter.

What is the phase of water at 1.0 atm and 50°C?

Gnom [1K]

Water vapor is on delivered in your room number so you don’t need it for your phone number though you need

6 0

2 years ago

What is the stock and classical name for CrBr3?

madam [21]

Answer:

The stock name (CrBr3) : <em>chromium(III) sulfide</em>

The classical name (CrBr3) : <em>chromic bromide</em>

3 0

4 years ago

The production of ammonia (NH3) under standard conditions at 25°C is represented by the following thermochemical equation. N2(g)

asambeis [7]

Answer:

44,901 kilo Joule heat is released when $1.663\times 10^4 g$ grams of ammonia is produced.

Explanation:

Moles of ammonia gas produced :

$\frac{1.663\times 10^4 g}{17 g/mol}=978.235 mol$

According to reaction, when 2 moles of ammonia are produced 9.18 kilo joules of energy is also released.

So, When 978.235 moles of ammonia gas is produced the energy released will be:

$\frac{-91.8 kJ}{2}\times 978.235 mol=-44,900.98 kJ\approx -44,901 kJ$

(negative sign indicates that energy is released as heat)

44,901 kilo Joule heat is released when $1.663\times 10^4 g$ grams of ammonia is produced.

7 0

3 years ago

Nitric oxide (NO) from car exhaust is a primary air pollutant. Calculate the equilibrium constant for the reaction

viva [34]

This problem is asking for the equilibrium constant at two different temperatures by describing the chemical equilibrium between gaseous nitrogen, oxygen and nitrogen monoxide at 25 °C and 1496 °C as the room temperature and the typical temperature inside the cylinders of a car's engine respectively:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Thus, the calculated equilibrium constants turned out to be 6.19x10⁻³¹ and 9.87x10⁻⁵ at the aforementioned temperatures, respectively, according to the following work:

There is a relationship between the Gibbs free energy, enthalpy and entropy of the reaction, which leads to the equilibrium constant as shown below:

$\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=-RT ln(K)$

Which means we can calculate the enthalpy and entropy of reaction and subsequently the Gibbs free energy and equilibrium constant. In such a way, we calculate these two as follows, according to the enthalpies of formation and standard entropies of N₂(g), O₂(g) and NO(g) since these are assumed constant along the temperature range:

$\Delta _rH=2*90.25 kJ/mol - (0 kJ/mol+0 kJ/mol)=180.5kJ/mol\\\\\Delta _rS=2*(0.211 kJ/mol*K)-(0.192kJ/mol*K+0.205kJ/mol*K)=0.025kJ/mol*K$

Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:

$\Delta _rG_{25\°C}=180.5-(25+298.15)*0.025=172.42kJ/mol\\\\\Delta _rG_{1496\°C}=180.5-(1496+298.15)*0.025=135.65kJ/mol$

And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:

$K=exp(-\frac{\Delta _rG}{RT} )\\\\K_{25\°C}=exp[-\frac{172420 J/mol}{(8.3145\frac{J}{mol*K})(298.15K)} ]=6.19x10^{-31}\\\\K_{1496\°C}=exp[-\frac{135650J/mol}{(8.3145\frac{J}{mol*K})(1769K)} ]=9.87x10^{-5}$

Learn more:

(Gibbs free energy) brainly.com/question/15213613

4 0

3 years ago

How many liters of hydrogen gas is produced from 3.712 g of magnesium with 104.2ml of 1.385 mol/L HCL (aq) at SATP? Please show

Licemer1 [7]

Answer:

$V=1.61L$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Next, we compute the reacting moles of each reactants:

$n_{Mg}=3.712gMg*\frac{1molMg}{24.305 gMg}=0.153molMg$

$n_{HCl}=1.385\frac{molHCl}{L}*0.1042L=0.144molHCl$

Then, as magnesium and hydrohloric acid are in a 1:2 molar ratio 0.153 moles of magnesium will completely react with 0.306 moles of hydrochloric acid yet we only have 0.144 moles, therefore, limiting reactant is hydrochloric acid. Thus, we compute the produced moles of hydrogen:

$n_{H_2}=0.144molHCl*\frac{1molH_2}{2molHCl} =0.072molH_2$

Finally, we use the ideal gas equation with T=298K and 1atm (STP conditions) to compute the liters of hydrogen gas:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{0.072mol*0.082\frac{atm*L}{mol*K}*273K}{1atm}\\ \\V=1.61L$

Best regards.

3 0

3 years ago