If ice is warmed and becomes a liquid, the process is endothermic.
The process requires heat in order to proceed. If ice stays in a very cold place, it will not melt unless it's heated. If ice is placed outside where it melts on its own, it gets the heat from the surroundings.
Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
Answer:
M HCl sln = 12.0785 M
Explanation:
- molarity (M) [=] mol/L
- %mm = ((mass compound)/(mass sln))*100
∴ mass sln = 100.0 g
∴ δ sln = 1.19 g/mL
∴ % m/m = 37 %
⇒ 37 % =((mass HCl/mass sln))*100
⇒ 0.37 = mass HCl / 100.0 g
⇒ 37 g = mass HCl
∴ molar mass HCl = 36.46 g/mol
⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol
⇒ volume sln = (100 g sln)*(mL/1.19 g) = 84.034 mL = 0.084034 L
⇒ M HClsln = 1.015 mol/0.084034 L
⇒ M HCl sln = 12.0785 M
0.003 moles of NaOH was used in the titration.
<h3>What is titration?</h3>
The concentration of an identified analyte can be found using a simple laboratory technique called titration. As a standard solution with a given concentration and volume, a reagent known as the titrant or titrator is created.
By using a solution with a known concentration to measure the concentration of an unknown solution, this process is known as titration. To a known volume of the analyte (the unknown solution), the titrant (the known solution) is typically added from a buret until the reaction is finished. To ascertain the unknown concentration of an identifiable analyte, titration, commonly referred to as titrimetry, is a widely used quantitative laboratory analytical technique (Medwick and Kirschner, 2010). Volume measurements are a crucial component of titration
Concentration in mol/dm3 =
Amount of solution mol
= concentration in mol/dm3 × volume in dm3
Amount of sodium hydroxide
= 0.100 × 0.0250
= 0.00250 mol
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