Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:

First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?

= 25000 g
Substitute the values into the formula:

= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.
Answer:
Relating to orbit or orbits
Explanation:
Drill cores from the ocean floor were dated and found to be very young compared to the age of the earth. This means the crust had to be formed recently, which can be explained by creation of crust at a spreading center.
A the type of soil the tulip receive- with or without fertilizer
Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [HA]
-------- (1)
Solution:
Convert Ka into pKa,
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.