Answer:
0.00500M of Na₂C₂O₄
Explanation:
<em>When are dissolved in 150 mL of 1.0 M H2SO4.</em>
<em />
We can solve this problem finding molarity of sodium oxalate: That is, moles of Na2C2O4 per liter of solution. Thus, we need to convert the 0.1005g to moles using molar mass of sodium oxalate (134g/mol) and dividing in the 0.150L of the solution:
0.1005g * (1mol / 134g) = 7.5x10⁻⁴ moles of Na₂C₂O₄
In 0.150L:
7.5x10⁻⁴ moles of Na₂C₂O₄ / 0.150L =
<h3>0.00500M of Na₂C₂O₄</h3>
Answer:
-4.59°C
Explanation:
Let's see the formula for freezing point depression.
ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Kf = Freezing constant. For water if 1.86°C/m
m = molality (moles of solute in 1kg of solvent)
i = Van't Hoff factor.
0°C - Freezing T° of solution = 1.86°C /m . 1.30m . 1.9
Freezing T° of solution = - (1.86°C /m . 1.30m . 1.9)
Freezing T° of solution = - (1.86°C/m . 1.30m . 1.9) → -4.59°C
NaCl → Na⁺ + Cl⁻
ANSWER:
Group 18:
Elements are
-helium(He)
-neon (Ne)
-argon (Ar)
-krypton (Kr)
-xenon (Xe)
-radon (Rn)
-oganesson (Og)
Answer:
The frequencies of the two lines are:
a) 
b)
When we heat rubidium compound we will see red color.
Explanation:
c = speed of light
= wavelength of light
a) Frequency of the light when wavelength is equal to 
This frequency corresponds to red light
b) Frequency of the light when wavelength is equal to 
This frequency corresponds to violet light
When we heat rubidium compound we will see red color.
