Given :
A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
To Find :
The volume of the gas after it is heated.
Solution :
Since, their is no information about pressure in the question statement let us assume that pressure is constant.
Now, we know by ideal gas equation at constant pressure :
Hence, this is the required solution.
The two main factors the temperature of seawater are density and the salinity of the water.
A. the wax is a both; 1. physical change-solid to liquid.
2. chemical change- burned to CO2 + H20 + heat + carbon as seen as black on the rod
b. the wick is neither; the wick does not change, just provides conduit for wax to flame
c. the glass rod is physical change; the carbon is only deported
HOPE THIS HELPS, IVE ALSO LEARNING BEEN LEARNING THIS RECENTLY
The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.
<u>Explanation:</u>
Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.
As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.
Hence, 
Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
