I remember learning this last y’all i jus don’t remember it might be A or C
Answer: Physical change : tearing of paper, fixing of wtaer
Chemical change: rusting of iron , electrolysis of water, Rancidification
Explanation:
Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.
Example: tearing of paper, fixing of wtaer
Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.
Example: rusting of iron , electrolysis of water, Rancidification
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
<u>Answer:</u> The solubility of
in water is 
<u>Explanation:</u>
The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

3s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5E3%5BPO_4%5E%7B3-%7D%5D%5E2)
We are given:

Putting values in above equation, we get:

Hence, the solubility of
in water is 
Answer:
33.33% = 33%
Explanation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
1 mole of MCO3 will produce → 1 mole of CO2
We need to get the number of mole of CO2:
and when we have 0.22 g of CO2, so number of mole = mass / molar mass
Moles = 0.22 g / 44 g/mol = 0.005 mole
Moles of Mg = moles of CO2 = 0.005 mole
Mass of Mg = moles * molar mass
= 0.005 * 84 /mol = 0.42 g
Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100
=33.33 %