Complete Question:
This diagram shows a marble with a mass of 3.8 grams (g) that was placed into 10 milliliters (mL) of water. Using the formula V M D = , what is the density of the marble?
(See attachment for full diagram)
Answer:
1.27 g/cm³
Explanation:
First, find the volume of rock:
Volume of rock = volume of water after rock was placed - volume of water before rock was placed
Volume of rock = 13 - 10 = 3ml
Density of rock = grams of rock per 1 cm³
Note: 1 ml = 1 cm³
Let x represent amount of rock per 1 cm³
Thus,
3.8g = 3 cm³
x = 1 cm³
Cross multiply
1*3.8 = 3*x
3.8 = 3x
3.8/3 = 3x/3
1.27 = x
Density of rock = 1.27 g/cm³
C. TO GATHER AS MUCH INFORMATION ON THE TEXT AS POSSIBLE SO YOU CAN BETTER UNDERSTAND THE TEXT WHEN YOU READ
The symbols for three different cations with 14 protons are Si²⁺, Si⁴⁺, and Si³⁺.
An element with 14 protons must be <em>silicon, S</em>i.
Its electron configuration is [Ne] 3s²3p², so it could lose up to four valence electrons.
The most likely cations are Si²⁺, Si⁴⁺, and Si³⁺ (or Si⁺).
A nuclear reactionIt’s like another particle with the release of energy
The reaction of acetic acid with sodium hydroxide is:
The ratio of A-/HA is calculated as follows:
According to Henderson Hasslebach equation:
![[A-]/[HA] = 10^p^H^-^p^K^_a](https://tex.z-dn.net/?f=%5BA-%5D%2F%5BHA%5D%20%3D%2010%5Ep%5EH%5E-%5Ep%5EK%5E_a)
The total concentration of HA and A- = 2.0 L * 0.25 M = 0.5 mol.
![[ A- ]+ [ HA ]= 0.5](https://tex.z-dn.net/?f=%5B%20A-%20%5D%2B%20%5B%20HA%20%5D%3D%200.5)
![[ A^- ] = 0.5 – [ HA]](https://tex.z-dn.net/?f=%5B%20A%5E-%20%5D%20%3D%200.5%20%E2%80%93%20%5B%20HA%5D)
[ HA] = 0.156 mol and [ A- ]= 0.343 mol
Total of 0.5 moles of acetic acid is required:
HA = 30.0 g acetic acid
Conversion of 0.343 moles of the acetic acid to acetate can be performed by adding NaOH
= 343 mL
Thus, 343 mL of 1M NaOH is required
