Answer:
39.998
Explanation:
Hello, have a periodic table handy!
Molar mass is the molecular weight of a compound; sum of all of the mass in grams (from the periodic table) from each and specific atom.
In the periodic table, the molecular weight of an atom is usually shown on the bottom of the element.
In NaOH, there are three atoms in that compound. Na (sodium), O (oxygen), and H (hydrogen).
For Na, the molecular weight is 22.99
For O, the molecular weight is 16.00
For H, the molecular weight is 1.008.
Add all of those values up together to get the molar mass of NaOH being 39.998
We assume that the given gas, NO2, is an ideal gas such that we will be able to use the equation,
n/V = P/RT
where n is the number of moles, V is volume, P is pressure, R is gas constant, and T is temperature in kelvin. Substituting the known values,
n/V = (1 atm) / (37 + 273)(0.0821 L.atm/mol.K)
n/V = 0.039 mol/L
Converting this to density by multiplying the value with the molar mass,
density = (0.039 mol/L) x (46 g/mol) = 1.8 g/L
Answer:
the answer is second option. with 2 models
Explanation:
2H2 + O2 gives 2H2O
so there's two models
Answer:
THE FREEZING POINT OF A WATER SOLUTION OF FRUCTOSE MADE BY DISSOLVING 92 g OF FRUCTOSE IN 202g OF WATER IS -4.70 ◦C
Explanation:
To calculate the freezing point of a water solution of fructose,
1. calculate the molar mass of Fructose
( 12 * 6 + 1*12 + 16*6) =72 + 12 +96 = 180 g/mol
2. calculate the number of moles of fructose in the solution
number of moles = mass / molar mass
n = 92 g / 180 g/mol
n = 0.511 moles.
3. calculate the molarity of the solution
molarity = moles / mass of water in kg
molarity = 0.5111 / 202 g /1000 g
molarity = 0.5111 / 0.202
molarity = 2.529 M
4. calculate the change in the freezing point of pure solvent and solution ΔTf
ΔTf = Kf * molarity of the solute
Kf = 1.86 ◦C/m for water
ΔTf = 1.86 * 2.529
ΔTf = 4.70 C
5. the freezing point is therefore
0.00 ◦C - 4.70 ◦C = -4.70 ◦C
Answer: 0.036 J/g°C
Explanation:
The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Given that,
Q = 3.42 Kilojoules
[Convert 3.42 kilojoules to joules
If 1 kilojoule = 1000 joules
3.42 kilojoules = 3.42 x 1000 = 3420J]
Mass = 2.508Kg
[Convert 2.508 kg to grams
If 1 kg = 1000 grams
2.508kg = 2.508 x 1000 = 2508g]
C = ? (let unknown value be Z)
Φ = (Final temperature - Initial temperature)
= 42.061°C - 4.051°C
= 38.01°C
Apply the formula, Q = MCΦ
3420J = 2508g x Z x 38.01°C
3420J = 95329.08g•°C x Z
Z = (3420J / 95329.08g•°C)
Z = 0.03588 J/g°C
Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C
Thus, the specific heat of the substance is 0.036 J/g°C
