Answer:
For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%
Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver
12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:
38.1-g/43.2-g x 100% = 88.2%
Explanation:
Answer:
Lithium
Explanation:
it is the only metal listed.
Answer:
The answer to your question is it is not at equilibrium, it will move to the products.
Explanation:
Data
Keq = 2400
Volume = 1 L
moles of NO = 0.024
moles of N₂ = 2
moles of O₂ = 2.6
Process
1.- Determine the concentration of reactants and products
[NO] = 0.024 / 1 = 0.024
[N₂] = 2/1 = 2
[O₂] = 2.6/ 1= 2.6
2.- Balanced chemical reaction
N₂ + O₂ ⇒ 2NO
3.- Write the equation for the equilibrium of this reaction
Keq = [NO]²/[N₂][O₂]
- Substitution
Keq = [0.024]² / [2][2.6]
-Simplification
Keq = 0.000576 / 5.2
-Result
Keq = 1.11 x 10⁻⁴
Conclusion
It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-
1.11 x 10⁻⁴ < 2400
Answer:
Earth's average albedo is about 0.3. In other words, about 30 percent of incoming solar radiation is reflected back into space and 70 percent is absorbed. A sensor aboard NASA's Terra satellite is now collecting detailed measurements of how much sunlight the earth's surface reflects back up into the atmosphere.
Explanation:
hope this helps :)
Answer:
Joan should measure 0.625 mL of 2,12 M HCl solution to create her desired solution.
Explanation:
Molarity of HCl Joan has access to = ![M_1=2.12 M](https://tex.z-dn.net/?f=M_1%3D2.12%20M)
Volume of 2.12 M of HCl Joan use = ![V_1=?](https://tex.z-dn.net/?f=V_1%3D%3F)
Molarity of HCl Joan desired = ![M_2=0.053 M](https://tex.z-dn.net/?f=M_2%3D0.053%20M)
Volume of 0.053 M of HCl Joan can prepare = ![V_2=25 mL](https://tex.z-dn.net/?f=V_2%3D25%20mL)
![M_1V_1=M_2V_2](https://tex.z-dn.net/?f=M_1V_1%3DM_2V_2)
![V_1=\frac{M_2V_2}{M_1}](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7BM_2V_2%7D%7BM_1%7D)
![=\frac{0.053 M\times 25 mL}{2.12 M}=0.625 mL](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.053%20M%5Ctimes%2025%20mL%7D%7B2.12%20M%7D%3D0.625%20mL)
Joan should measure 0.625 mL of 2,12 M HCl solution to create her desired solution.