Answer:
Tienes que poner los elementos de las clasificaciones del agua y sus nombres.
Explanation:
no se explicar pero busca las clasificaciones del agua y las pones nada mas.
Phosphoric acid is a weak acid, while rubidium hydroxide is a strong acid.
H₃PO₄ + RbOH --> Rb₃PO₄ + H₂O
We get Rb₃PO₄ because PO₄ has a charge of 3-, that is PO₄³⁻. Rb has a charge of 1+. You give the subscript of one the charge of the other as this is an ionic compound. So you end up with Rb₃PO₄, a neutral compound.
Now let's balance the equation:
H₃PO₄ + 3RbOH --> Rb₃PO₄ + 3H₂O
<u>Answer:</u> The molarity of sulfuric acid is 0.0946 M
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![H_2SO_4](https://tex.z-dn.net/?f=H_2SO_4)
are the n-factor, molarity and volume of base which is LiOH
We are given:
![n_1=2\\M_1=?M\\V_1=52.87mL\\n_2=1\\M_2=1.25M\\V_2=8.00mL](https://tex.z-dn.net/?f=n_1%3D2%5C%5CM_1%3D%3FM%5C%5CV_1%3D52.87mL%5C%5Cn_2%3D1%5C%5CM_2%3D1.25M%5C%5CV_2%3D8.00mL)
Putting values in above equation, we get:
![2\times M_1\times 52.87=1\times 1.25\times 8.00\\\\M_1=\frac{1\times 1.25\times 8.00}{2\times 52.87}=0.0946M](https://tex.z-dn.net/?f=2%5Ctimes%20M_1%5Ctimes%2052.87%3D1%5Ctimes%201.25%5Ctimes%208.00%5C%5C%5C%5CM_1%3D%5Cfrac%7B1%5Ctimes%201.25%5Ctimes%208.00%7D%7B2%5Ctimes%2052.87%7D%3D0.0946M)
Hence, the molarity of sulfuric acid is 0.0946 M
First statement
Fifth statement
Sixth statement