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skad [1K]
2 years ago
11

Hydrogen, when combined with nitrogen in a chemical reaction, forms ammonia. Which type of chemical reaction is this an example

of?
Chemistry
1 answer:
larisa86 [58]2 years ago
7 0

Answer: Combination reaction

Explanation:

Hydrogen (H2) combines with nitrogen (N2) to form a single compound, ammonia (NH3) in a combination reaction.

The chemical equation is as follows:

N2(g) + 3H2(g) --> 2NH3(g)

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Answer these please ASAP need help no idea how to do these
STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass =  g

Cl₂:

Number of moles = Mass / molar masa

Number of moles  = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol  = Mass / 2 g/mol

Mass =  16 g

P₄:

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles  = 1.6 g /48  g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles  = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles  = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

               Pb         :        O₂

                2          :         1

               0.03     :      1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass =  0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂   → 2PbO

   

6 0
2 years ago
4Hg(OH)2+ H3PO4 —> Hg3(PO4)2 + 5H20
Lorico [155]

Answer:

Hi. im an online tutor and i ca help you with all your assignments . check out our wesite https://toplivewriters.com

Explanation:

3 0
2 years ago
A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm
Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys)  =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

<u>Step 1</u>: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

<u>(a) reversibly</u>

<u>Step 2:</u> Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

<u>Step 3:</u> Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

<u>Step 4:</u> Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)  

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

<u>(b) against a constant external pressure of 1.00 atm</u>

<u>Step 1</u>: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

   ⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

     ⇒ Vf = (1*0.08206 *300)/1

      ⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

<u>Step 2:</u> Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

<u>Step 3: </u>Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0
3 years ago
Is a landslide an example of erosion or weathering
Shtirlitz [24]

Answer:

It is a example of erosion

Explanation: because pieces of rock erode making it unstable a causes a landslide

7 0
3 years ago
When salt dissolves in water, the water molecules are attracted by dissolves salt particles. This attraction is called ?
Elza [17]

When salt dissolves in water, the water molecules are attracted by dissolves salt particles. This attraction is called dissociation.

Hope this helps you. Do mark me as brainliest.

4 0
2 years ago
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