Answer:
B. a H+ ion is the answer dear.
Explanation:
ㅗㅐㅔㄷ ㅅㅗㅑㄴ ㅗㄷㅣㅔ ㅛㅐㅕ.
Answer:
5.37 × 10⁻⁴ mol/L
Explanation:
<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.00154 mol/L
- Initial volume (V₁): 230. mL
- Final concentration (C₂): ?
- Final volume (V₂): 660. mL
Step 2: Calculate the concentration of the final solution
We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁ / V₂
C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L
Answer:
See Explanation
Explanation:
An endothermic reaction is one in which energy is absorbed and the change in enthalpy for the reaction is positive.
If we look at the reaction 2HgO + 45kcal ----> 2Hg + O2; we will notice that
i) 45kcal of energy was taken in (absorbed) for the reaction to occur
ii) The value of the reaction enthalpy is positive
For these two reasons, the reaction is an endothermic reaction as written.
The purpose of a chemical equation is to relate the amounts of reactants to the amounts of products based on the rate each is consumed. In this problem, one mole of sulfuric acid is consumed along with two moles of sodium cyanide to produce two moles of hydrocyanic acid and one mole of sodium sulfate. The relationship between sodium cyanide and sodium sulfate is 2:1, meaning that two moles of NaCN is required to produce one mole of sodium sulfate.
To produce 4.2 moles of sodium sulfate, two times this amount of NaCN is required. This means that you would need 8.4 moles of sodium cyanide.
Hope this helps!
Answer:
(1) addition of HBr to 2-methyl-2-pentene
Explanation:
In this case, we will have the formation of a <u>carbocation</u> for each molecule. For molecule 1 we will have a <u>tertiary carbocation</u> and for molecule 2 we will have a <u>secondary carbocation</u>.
Therefore the <u>most stable carbocation</u> is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)
