Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Answer:
the amount of neutrons would be the difference in the three isotopes
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−
Assuming all the Cr is contained in the BaCrO4 at the end.
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
I don't see the options for an answer, so here is a list of all of the transition metals lol
- <em>Scandium</em>
- <em>Titanium</em>
- <em>Vanadium</em>
- <em>Chromium</em>
- <em>Manganese</em>
- <em>Iron</em>
- <em>Cobalt</em>
- <em>Nickel</em>
- <em>Copper</em>
- <em>Zinc</em>
- <em>Yttrium</em>
- <em>Zirconium</em>
- <em>Niobium</em>
- <em>Molybdenum</em>
- <em>Technetium</em>
- <em>Ruthenium</em>
- <em>Rhodium</em>
- <em>Palladium</em>
- <em>Silver</em>
- <em>Cadmium</em>
- <em>Lanthanum</em>
- <em>Hafnium</em>
- <em>Tantalum</em>
- <em>Tungsten</em>
- <em>Rhenium</em>
- <em>Osmium</em>
- <em>Iridium</em>
- <em>Platinum</em>
- <em>Gold</em>
- <em>Mercury</em>
- <em>Actinium</em>
- <em>Rutherfordium</em>
- <em>Dubnium</em>
- <em>Seaborgium</em>
- <em>Bohrium</em>
- <em>Hassium</em>
- <em>Meitnerium</em>
- <em>Darmstadtium</em>
- <em>Roentgenium</em>
- <em>Copernicium p</em>
Answer:skin cream
Explanation:
the independent variable is the variable that is always changed or manipulated. in this case the skin cream is the independent variable.
