Increase at the temperature
Answer:
The answer to your question is letter A.
Explanation:
Isomers are molecules that have the same molecular formula but have a different structure. The molecule from which are looking an isomer has 5 carbons and 1 double bond. Then we need to look for another molecule with these components.
A.- This molecule has 5 carbons and 1 double bond, This structure is an isomer of the first one.
B.- This molecule has 3 carbons and 1 double bond, it's not an isomer of the first structure.
C. This molecule has 4 carbons and 1 triple bonds, it's not an isomer of the first structure.
D. This molecule has 5 carbons but it doesn't have any double bond, then it's not an isomer of the first structure.
Answer:
i) Dilute hydrochloric acid will react with Ammonia to form ammonia salt.
ii) dilute hydrochloric acid will react with soduim hydroxide to form sodium chloride and water
iii) dilute hydrochloric acid will react with calcuim carbonate to form Calcium chloride, Carbon dioxide and water.
Explanation:
CHEMICAL EQUATIONS :
I) HCL + NH3 = NH4Cl
ii) HCL+ NaOH = NaCl + H2O
iii) HCL + CaCo = CaCl2 + CO2 + H2O
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
