Answer:
The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose
Option d, b, a and c
Explanation:
Colligative property: Freezing point depression
The formula is: ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.
Option d, which is glucose is non electrolyte so the i = 1
a. NaI → Na⁺ + I⁻ i =2
b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3
c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4
Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.
Answer:
11.0 L
Explanation:
The equation for this reaction is given as;
2H2 + O2 --> 2H2O
2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O
At STP;
1 mol = 22.4 L
This means;
44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O
In this reaction, the limiting reactant is H2 as O2 is in excess.
The relationship between H2 and H2O;
44.8 L = 44.8 L
11.0 L would produce x
Solving for x;
x = 11 * 44.8 / 44.8
x = 11.0 L
Answer:
MOLAR MASS = 32 g/mol
Explanation:
Condition of standard temperature and pressure(STP) are as follow:
Temperature = 273 K
Pressure = 1 atm (or 100000 Pa)
Here atm is atmosphere and Pa is Pascal
STP conditions arte used for measuring gas density and volume using Ideal Gas Law.Here 1 mole of ideal gas occupies 22.4 L of volume.
According toi Ideal Gas Equation :
PV = nRT
where P = pressure, n= number of moles, V = volume ,R= Ideal Gas Constant and T= temperature
From question:
V=280 ml = 0.28 L
P = 1 atm
R=0.08205 L atm/K mol
T=273 K
Putting values in above formula :
n = 0.0125 moles
Now 
given mass = 0.4 g (given)
On solving we get:
Molar mass = 32 g/mol
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Water is an amphoteric compound. This means it could be a base or an acid, depending on the substance it is to be reacted with. In this case, water is a base because HF is an acid. Now, if the reactant is an acid, its form after the reaction is called the conjugate pair. Since HF became F⁻, <em>the acid-conjugate base pair is: HF and F⁻.</em>
