Answer:
1. Hydrogen ions; acidic
2. Alkali; hydroxide ions; alkaline
3a. Sulfuric acid --> 2 Hydrogen ions + sulfate ion
H₂SO₄ --> 2H+ + SO₄²-
3b. Sodium hydroxide --> Sodium ion + Hydroxide ion
NaOH --> Na+ + OH-
Explanation:
1. Sulfuric acid releases hydrogen ions in solution. This makes the solution acidic.
Acids produce hydrogen ions when dissolved in aqueous solutions.
2. Sodium hydroxide is an alkali. It releases hydroxide ions in solution. This makes the solution alkaline.
Alkalis are soluble bases that produce hydroxide ions in solution.
3a. Sulfuric acid --> 2 Hydrogen ions + sulfate ions
H₂SO₄ --> 2H+ + SO₄²-
The equation above is for the ionization of sulfuric acid
b. Sodium hydroxide --> Sodium ion + Hydroxide ion
NaOH --> Na+ + OH-
The equation above is for the ionization of sodium hydroxide
Answer:
glucose is produced as a sort of food for the plant
Answer: The standard enthalpy of formation of this isomer of 
is -210.9 kJ
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of formation of 
.
![\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CO_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%2Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}](https://tex.z-dn.net/?f=-511.3kJ%2Fmol%3D%5B%288%5Ctimes%20-393.5%29%2B%289%5Ctimes%20-241.8%29%5D-%5B%28%5Cfrac%7B25%7D%7B2%7D%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D)
Refraction is the answer
Gggggt gg for the ggv. Gv. Gc. Gv. CFC. F
<h3>
Answer:</h3>
318.405 g
<h3>
Explanation:</h3>
We are given;
- Volume of K₃PO₄ solution to be prepared as 750.0 mL or 0.75 L
- Molarity of the solution to be prepared as 2.00 M
We are required to determine the mass of K₃PO₄ to be measured.
<h3>Step 1: Determine the number of moles of K₃PO₄</h3>
Molarity = Moles ÷ Volume
Rearranging the formula;
Number of moles = Molarity × Volume
Therefore;
Moles of K₃PO₄ = 2.00 M × 0.75 L
= 1.5 moles
<h3>Step 2: Determine the mass of K₃PO₄ to be measured</h3>
Mass = Number of moles × Molar mass
Molar mass of K₃PO₄ = 212.27 g/mol
Therefore;
Mass of K₃PO₄ = 1.5 moles × 212.27 g/mol
= 318.405 g
Therefore, the mass of K₃PO₄ that should be weighed is 318.405 g
