Given:
density of water is <span>approximately 1.0 g/mL
block of material
</span><span>volume of 1.2 x 10^4 in^3
and weighing 350 lb
</span>
Required:
Which will float
Solution:
D = M/V where D is density, M is mass and V is volume
convert <span>1.2 x 10^4 in^3 into mL
</span><span>1.2 x 10^4 in^3 (16.39 mL/ 1 in3) = 196,980 mL
</span>
convert <span>350 lb into grams
350 lb (454 grams/1 lb) = 158,900 grams
D = M/V = </span>158,900 grams/ <span>196,980 mL
D = 0.807 g/mL
The block will float in water because it is less dense than water</span>
<span>when a gas is collected over water at 30.0 deg C , the H2O vapor pressure is 32 Torr
Part A:What is the partial pressure of the hydrogen gas collected in this way?
at a total pressure of 732 mm Hg. - 32 Torr H2O vapor = 700. Torr H2
your answer is
answer is 700 mm Hg.
Part B:
If the total volume of gas collected is 724 mL, what mass of hydrogen gas is collected?
find moles
PV = nRT
(700.Torr)(0.724 Litres) = n (62.36 Torr-Litres/mol-K) (303 Kelvin)
n = 0.02682 moles of H2
using molar mass, find moles:
( 0.02682 moles of H2) (2.016 grams H2 / mol) = 0.05407 grams of H2
your answer rounded to 3 sig figs is
0.0541 grams of H2</span>
Answer:
<h2>Answer⤵</h2><h3><u>The number in a mole, Avogadro's number, is related to the relative sizes of the atomic mass unit and gram mass units. Whereas one hydrogen atom has a mass of approximately 1 u, 1 mol of H atoms has a mass of approximately 1 gram.</u></h3>
Answer:
0.1
Explanation:
Using Henderson-Hasselbalch equation
pH = pKa + log ( base / acid )
pH= 9.0 , pKa = 10.0
9.0 = 10 + log ( base / acid )
9 - 10 = log ( base / acid )
10⁻¹ = ( base / acid )
( base / acid ) = 0.1