Answer:
0.2598 M
Explanation:
Molarity is mol/L, so we have to convert the grams to moles and the mL to L. To convert between grams and moles you need the molar mass of the compound, which is 36.46g/mol.


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Round to the lowest number of significant figures = 0.2598 M
Ice melts when heat energy causes the molecules to move faster, breaking the hydrogen bonds between molecules to form liquid water.
This question requires the knowledge of density.
The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³
Density = Mass / Volume
By applying ethyl alcohol,
789 kg m⁻³ = Mass / 0.9 m³
Mass = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.
Then by applying water,
1000 kg m⁻³ = 710.1 kg / Volume
Volume = 0.7101 m³
= 0.7 m³
hence the equal water volume is 0.7 m³
Answer:
V = 81.14 L
Explanation:
Given data:
Volume of gas = ?
Number of moles = 3.30 mol
Temperature of gas = 25°C
Pressure of gas = 0.995 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
25+273 = 298 K
now we will put the values in formula:
V = 3.30 mol 0.0821 atm.L/ mol.K 298 K / 0.995 atm
V = 80.74 L. atm / 0.995 atm
V = 81.14 L
Answer:
Reducing sugars are absent
Explanation:
Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.
Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.