How much heat is transferred when 7.19 grams of H2 reacts with excess nitrogen, according to the following equation: N2(g) + 3 H2 (g) --> 2 NH3 (g) \DeltaΔH = +46.2 kJ

Answers

Answer 1

Answer:

Answer:

$Q=54.8kJ$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible for us to realize that the 46.2 kJ of energy are given per mole of reaction, which are related to 3 moles of hydrogen; Thus, we can calculate the energy per mole of hydrogen as shown below:

$\Delta H=(46.2kJ)/(mol) *(1mol)/(3molH_2)\n\n \Delta H=15.4(kJ)/(molH_2)$

Now, to calculate the total energy, we convert the grams to moles of hydrogen as shown below:

$Q=7.19gH_2*(1molH_2)/(2.02gmolH_2)*15.4(kJ)/(molH_2) \n\nQ=54.8kJ$

Best regards!

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Answers

Answer:

Each nerve cell is connected to the brain by electrical signals

Explanation:

Answer:

they said it above me

Explanation:

Write the expression for the equilibrium constant Kp for the following reaction. Enclose pressures in parentheses and do NOT write the chemical formula as a subscript. For example, enter (PNH3 )2 as (P NH3)2. If either the numerator or denominator is 1, please enter 1 2 MoO3(s) ↔ 2 MoO2(s) + O2(g)

Answers

The expression for the equilibrium constant Kp for the following reaction is $K_p = (po_2)$

Equilibrium constant:

It refers to the ratio of the concentration of products to the concentration of reactants where each raised to the power of their stoichiometric ratios. It is expressed as K.

Since the given expression is

MoO3(s) ↔ 2 MoO2(s) + O2(g)

So, the above expression should be considered for constant kp

Learn more about reaction here; brainly.com/question/24185208

Answer: $K_p={(p_(O_2))}$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K$

$K_p$ is the constant of a certain reaction at equilibrium for gaseous reactants and products.

For the given chemical reaction:

$MoO_3(s)\rightleftharpoons 2MoO_2(s)+O_2(g)$

The expression of for above equation follows:

$K_p={(p_(O_2))}$

As solids do not exert pressure, $MoO_3$ and $MoO_2$ are not involved.

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255?

Answers

Answer is 231.8 K

Explanation
We can use Charle's law to solve this problem.

Charle's law says"at a constantpressure, the volume of a fixed amount of gas is directly proportional to itsabsolute temperature".

V α T

Where V is the volume and T is the temperaturein Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 1.00 L
T₁ =?
V₂ = 1.10 L
T₂ = 255 K (Since the unit of the given temperature is missing in the question, thought that the given temperature is in Kelvin)

By applying theformula,
1.00 L / T₁ = 1.10 L / 255 K
T₁ =(1.00 L / 1.10 L) x 255 K
T₁ = 231.8 K

Hence, the initial temperature is 231.8 K.

Assumption : the pressure of the gas is a constant.

Hello!

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255 ?

We have the following data:

V1 (initial volume) = 1.00 L

V2 (final volume) = 1.10 L

T1 (initial temperature) = ? (in Kelvin)

T2 (final temperature) = 255 K

According to the Law of Charles and Gay-Lussac in the study of gases, in an isobaric transformation, ie when a mass under pressure maintains its constant pressure, on the other hand, as the volume increases, the temperature increases and, if the volume decreases, the temperature decreases (directly proportional to temperature and volume) . We apply the data to the formula of isobaric transformation (Charles and Gay-Lussac), we will see:

$(V_1)/(T_1) = (V_2)/(T_2)$

$(1.00)/(T_1) = (1.10)/(255)$

multiply the means by the extremes

$1.10*T_1 = 1.00*255$

$1.10\:T_1 = 255$

$T_1 = (255)/(1.10)$

$\boxed{\boxed{T_1 \approx 231.8\:K}}\:\:\:\:\:\:\bf\red{\checkmark}$

Answer:

The initial temperature is approximately 231.8 Kelvin

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Which process is used to make lime (calcium oxide) from limestone (calcium carbonate)?

Answers

Answer:

Explanation:

Calcium oxide is fromed by the decomopostion of CaCO3 at high temperature.

CaCO3 ------> CaO +CO2

Hope this helps you

Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Explanation:

(a)

Initial number of moles of $NH_(3)$ and $O_(2)$ are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

$4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O$

The stoichiometric numbers are as follows.

$v_{NH_(3)}=-4$

$v_{O_(2)}=-5$

$v_(NO)=4$

$v_{H_(2)O}=6$

The total number of moles initially present -7

$\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

Initial number of moles of $H_(2)S$ and $O_(2)$ are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

$2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)$

The stoichiometric numbers are as follows.

$v_{H_(2)S}=-2$

$v_{O_(2)}=-3$

$v_{H_(2)O}=2$

$v_{SO_(2)=2$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

Initial number of moles of $NO_(2)$ , $NH_(3)$ and $N_(2)$ are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

$6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O$

The stoichiometric numbers are as follows.

$v_{NO_(2)}=-6$

$v_{NH_(3)}=-8$

$v_{N_(2)}=7$

$v_{H_(2)O}=12$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

Mole fraction of NH3: (2 - x)/(Total moles)
Mole fraction of O2: (5 - 5x/4)/(Total moles)
Mole fraction of NO: (4x)/(Total moles)
Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.