1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Naddika [18.5K]
4 years ago
6

According to the equation below, how many moles of PbO are required to generate 3.88×1023 nitrogen molecules?

Chemistry
1 answer:
saul85 [17]4 years ago
8 0

Answer:

1.935 mole

Explanation:

We'll begin by calculating the number of mole present in 3.88x10^23 molecules of nitrogen(N2). This can be obtained as follow:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 molecules. Therefore 1 mole of N2 contains 6.02x10^23 molecules.

Now if 1 mole of N2 contains 6.02x10^23 molecules,

Then Xmol of N2 will contain 3.88x10^23 molecules i.e

Xmol of N2 = (3.88x10^23)/6.02x10^23

Xmol of N2 = 0.645 mole

Now, we can obtain the number of moles of PbO required to generate 3.88x10^23 molecules (i.e 0.645 mole) of N2. This is illustrated below:

The equation for the reaction is given below:

3PbO + 2NH3 → 3Pb + N2 + 3H2O

From the balanced equation above, 3 moles of PbO produced 1 mole of N2.

Therefore, Xmol of PbO will produce 0.645 mole of N2 i.e

Xmol of PbO = 3 x 0.645

Xmol of PbO = 1.935 mole.

From the calculations made above,

1.935 mole of PbO will produce 3.88x10^23 molecules of nitrogen (N2).

You might be interested in
How many protons dose p have
Softa [21]

Answer:

i have to see the question

Explanation:

5 0
3 years ago
Read 2 more answers
A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g
svetoff [14.1K]

Answer:

The answer to your question is a) 51.07 g of NiBr₂   b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

mass of Br = 37.3 g

Balanced chemical reaction

                Ni  +  Br₂   ⇒   NiBr₂

Process

1.- Find the atomic mass of the reactants and the molar mass of the product

Ni = 59 g

Br = 79.9 x 2 = 159.8 g

NiBr₂ = 59 + 159.8 = 218.8 g

2.- Find the limiting reactant

theoretical yield  Ni/Br₂ = 59/159.8 = 0.369

experimental yield Ni/Br₂ = 67.8/37.3 = 1.81

The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

                    159.8 g of Br₂ --------------- 218.8 g of NiBr₂

                      37.3 g of Br₂ --------------  x

                          x = (37.3 x 218.8) / 159.8

                          x = 8161.24/159.8

                          x = 51.07 g of NiBr₂

4.- Find the excess reactant

The excess reactant is Nickel

                59 g of Ni ---------------- 159.8 g of Br₂

                  x               ----------------  37.3 g of Br₂

                            x = (37.3 x 59)/159.8

                            x = 2200.7/159.8

                            x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

                 = 54 g

5 0
3 years ago
What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o
Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

The formula can be written for the calculation of moles as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Thus,  

Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}

Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles

Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles

The chemical reaction taking place:

AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)

According to reaction stoichiometry:

<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}

Molarity\ of\ NaCl= 0.3374 M

<u>Thus, the concentration of NaCl = 0.3374 M</u>

6 0
3 years ago
In a pendulum, a continuous change occurs between kinetic energy and
Masja [62]
It should be potential energy!!!
7 0
3 years ago
Hey guys please help me
Anvisha [2.4K]

Answer:

Water will move out of the eggplant cells and the cells will shrink.

Explanation:

I found it on a Quizziz.

6 0
2 years ago
Other questions:
  • Calculate the heat needed to increase the temperature of 100. g water from 45.7 C to 103.5 C.
    13·2 answers
  • Find degenerate orbitals 3dxy, 2px, 4dz2, 3dyz, 3pz,3dxz, 4sm​
    13·1 answer
  • What volume of gold would be equal in mass to a piece of copper with a volume of 101 ml? the density of gold is 19.3 g/ml; the d
    10·2 answers
  • What is the molar mass of AlCl3?
    11·1 answer
  • What will be the the net ionic equations for the following ones:- a) AgNO3 + KCl b) Ni(NO3)2 + Na2S c) CaCl2 + Na2CO3. 2) Write
    5·1 answer
  • The heat transfer depicted in the image is MOST likely
    13·2 answers
  • 0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?
    9·1 answer
  • PLEASE HELP BRAINLIEST 4 BEST ANSWER
    10·2 answers
  • Help! please and thank youu
    12·2 answers
  • What is the mass of 8.03 mole of NH3? (NH3 is 17g/mol)
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!