Answer:

Explanation:
They gave us the masses of two reactants and asked us to determine the mass of the product.
This looks like a limiting reactant problem.
1. Assemble the information
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 239.27 32.00 207.2
2PbS + 3O₂ ⟶ 2Pb + 2SO₃
m/g: 2.54 1.88
2. Calculate the moles of each reactant

3. Calculate the moles of Pb from each reactant

4. Calculate the mass of Pb

Answer:
1: [H+] = 0.01 M
2: [H+] = 0.0001 M
3: [H+] = 0.0001 M
Explanation:
Step 1: data given
pH = -log[H+]
pH = pOH = 14
Step 2:
1. A solution with pH = 2.0
pH = 2
-log[H+] = 2.0
[H+] = 10^-2
[H+] = 0.01 M
2. A solution with pH = 4.0
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M
3. A solution with pOH = 10.0
pH = = 14 - 10 = 4
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M
Answer:
Explanation:
The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas.
The covalent bond between two atoms can be polar or nonpolar. If the atoms are equal, the bond will be nonpolar (since no atom attracts electrons more strongly). But, if the atoms are different, the bond will be polarized towards the most electronegative atom, because it will be the atom that attracts the electron pair with more force. Then it will be polar.
It can occur in a molecule that the bonds are polar and the molecule is nonpolar. This occurs because of the geometry of the molecule, which causes them to cancel the different equal polar bonds of the molecule.
In carbon tetrachloride the bonds are polar, but the tetrahedral geometry of the molecule causes all four dipoles to cancel out and the molecule to be apolar.
Answer:
(a) -0.00017 M/s;
(b) 0.00034 M/s
Explanation:
(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

Given the following reaction:

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:
![r = -\frac{\Delta [N_2O_5]}{2 \Delta t}](https://tex.z-dn.net/?f=r%20%3D%20-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D)
Reaction rate is also equal to the rate of formation of products divided by their coefficients:
![r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BO_2%5D%7D%7B%5CDelta%20t%7D)
Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

(b) Using the relationship derived previously, we know that:
![-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D)
Rate of appearance of nitrogen dioxide is given by:
![r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}](https://tex.z-dn.net/?f=r_%7BNO_2%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D)
Which is obtained from the equation:
![-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D)
If we multiply both sides by 4, that is:
![-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B4%20%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D)
This yields:
[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]