$d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}$

8 0

3 years ago

Potassium and silver replace each other in the following reaction: AgNO3 + KCl → KNO3 + AgCl. What type of reaction does this re

Sphinxa [80]

Notice how the K and Ag are both being swapped around.

Single Replacement:

A+BX → B+AX

Double Replacement:

AX+BY → BX + AY

7 0

3 years ago

) In another experiment, the student titrated 50.0mL of 0.100MHC2H3O2 with 0.100MNaOH(aq) . Calculate the pH of the solution at

andrew11 [14]

Answer:

Explanation:

moles of acetic acid = 500 x 10⁻³ x .1 M

= 5 X 10⁻³ M

.005 M

Moles of NaOH = .1 M

Moles of sodium acetate formed = .005 M

Moles of NaOH left = .095 M

pOH = 4.8 + log .005 / .095

= 4.8 -1.27875

= 3.52125

pH = 14 - 3.52125

= 10.48

6 0

3 years ago

_KOH + _H₂PO4 → _K₂PO4 + _H₂O​

_KOH + _H₂PO4 → _K₂PO4 + _H₂O