Answer:
Experience in star is very nice
Explanation:
Because i love stars
Answer:
![\large \boxed{\text{2.20 g Pb}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B2.20%20g%20Pb%7D%7D)
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
![\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20PbS%7D%20%3D%20%5Ctext%7B2.54%20g%20PbS%20%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20PbS%7D%7D%7B%5Ctext%7B239.27%20g%20PbS%7D%7D%20%3D%20%5Ctext%7B0.010%2062%20mol%20PbS%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20O%7D_%7B2%7D%20%3D%20%5Ctext%7B1.88%20g%20O%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20O%7D_%7B2%7D%7D%7B%5Ctext%7B32.00%20g%20O%7D_%7B2%7D%7D%20%3D%20%5Ctext%7B0.058%2075%20mol%20O%7D_%7B2%7D)
3. Calculate the moles of Pb from each reactant
![\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}](https://tex.z-dn.net/?f=%5Ctextbf%7BFrom%20PbS%3A%7D%5C%5C%5Ctext%7BMoles%20of%20Pb%7D%20%3D%20%20%5Ctext%7B0.010%2062%20mol%20PbS%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mol%20Pb%7D%7D%7B%5Ctext%7B2%20mol%20PbS%7D%7D%20%3D%20%5Ctext%7B0.010%2062%20mol%20Pb%7D%5C%5C%5C%5C%5Ctextbf%7BFrom%20O%7D_%7B2%7D%3A%5C%5C%5Ctext%7BMoles%20of%20Pb%7D%20%3D%5Ctext%7B0.058%2075%20mol%20O%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mol%20Pb%7D%7D%7B%5Ctext%7B3%20mol%20O%7D_%7B2%7D%7D%3D%20%5Ctext%7B0.039%2017%20mol%20%20Pb%7D%5C%5C%5C%5C%5Ctext%7BPbS%20is%20the%20%24%5Ctextbf%7Blimiting%20reactant%7D%24%20because%20it%20gives%20fewer%20moles%20of%20Pb%7D)
4. Calculate the mass of Pb
![\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20Pb%7D%20%3D%20%5Ctext%7B0.010%2062%20mol%20Pb%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B207.2%20g%20Pb%7D%7D%7B%5Ctext%7B1%20mol%20Pb%7D%7D%20%3D%20%5Ctextbf%7B2.20%20g%20Pb%7D%5C%5C%5C%5C%5Ctext%7BThe%20reaction%20produces%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B2.20%20g%20Pb%7D%7D%24%7D)
Answer: B
Explanation: the 3 in Na3PO4(s) belong to the sodium atom (Na). so in any of these equations, the 3 would have to be with Na.
A- the 3 is along w the PO4, which would make it part of that bond
C- there is no 3 at all for Na in this choice, making it incorrect
D- Again, the 3 is placed on the other half of the bond
The solution would be like this for this specific problem:
<span>Given:
</span>66.0 g of carbon monoxide
reaction 2 C + O2 → 2 CO
<span>mol e= mass / molar mass <span>
<span>mole of 2CO = 66.0g / (12.011 15.999)g / mol </span>
mole of 2CO = 2.36 (CO and C has a 1:1 mole ratio)
mole of 2CO = 2.36 -> mole of 1 CO = 2.36 / 2 = 1.18
<span>We got 2 moles of C, thus 1.18 x 2 = 2.36
So, we 2.36 </span>moles of carbon are needed to produce 66.0 g of carbon monoxide in the </span>reaction
2 C + O2 → 2 CO.</span>
<span>To add, Carbon nonmetallic
and tetravalent, thus, making four electrons available to form covalent
chemical bonds. </span>
Answer:
A
Explanation:
a compound is a pure substance and A tells that