M1V1 = M2V2
.200 (.025) = 1.60 X 10 -2 (V2)
V2 = .315 L
1.60 x 10-2 M in 315 mL
Answer:
The answer to your question is: letter D.
Explanation:
Double replacement reaction is when the cation of one reactant replaces the metal of the other reactant and vice versa.
a. 2SO2 + O2 —> 2S03 This is a combination reaction
b. Zn + Cu(NO3)2 → Zn(NO3)2 + Cu This is a single replacement reaction.
c. 2H2O2–> 2H2O + O2 This is a decomposition reaction
d . AgNO3 + NaCl → AgCl + NaNO3 This is a double replacement reaction
<u><em>Answer</em></u>
<u><em>Explanation</em></u>
- In periodic table, the elements have almost same properties are present in the same group. As Mg and Sr are present in group II-A, so both behave most likely to each other due to having same valence shell electrons as well.
- Si and Sn are present in group IV-A which have same behavior but different one from Mg due to different groups.
- S is present in group VI-A which show different properties from all others one especially from Mg.
Answer:
24x10³
Explanation:
2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)
The equilibrium constant for this reaction is:
Kc = ![\frac{[O_2]^3}{[CO_2]^2[H_2O]^4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BO_2%5D%5E3%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E4%7D)
The expression of [CH₃OH] is left out as it is a pure liquid.
Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:
- CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
- H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
- O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂
Then we calculate the concentrations:
- [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
- [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
- [O₂] = 0.0875 mol / 7.5 L = 0.0117 M
Finally we <u>calculate Kc</u>:
- Kc =
= 24x10³
