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
An Electrons movement is related
to it's amount of energy . In fact ,everything is related to it's amount of energy
[hope this helps]
Answer:
4.95
Explanation:
1.00 M H2A
1.38 m NaOH
Titration = 200.0 mL
Calculate moles of NaOH
=
= 0.46
calculate moles of H2A
=
= 0.667
therefore the moles of acid left = moles of H2A - moles of NaOH
= 0.667 - 0.46 = 0.207
pka = - log( ka )
= - log ( 2.5 * 10^-5 ) = 4.61
calculate PH after 100 ml of 1.38 M NaOH have been added
PH = pka + log ![(\frac{salt}{acid} )](https://tex.z-dn.net/?f=%28%5Cfrac%7Bsalt%7D%7Bacid%7D%20%29)
= 4.61 + log
= 4.95