<span>led to the discovery of many new sea floor features of smaller scales. Deep sea trenches and sea volcanoes.
If that helped please mark brainliest! <3
-Procklownyt</span>
Answer:
(a) 13.7 g.
(b) 28.91 g.
Explanation:
- molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.
∴ m = (no. of moles of solute)/(mass of water (kg))
<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>
<em />
<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>
∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).
m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).
∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.
<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>
∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).
m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).
∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.
Answer:
The pressure of the gas increased (if temperature remained constant).
The Boyle's law supports this observation.
Explanation:
The initial measurements of the gas are given as;
volume = 100 L
Pressure = 300 kpa
The second measurement is given as;
Volume = 75 L
The second reading implies that the volume of the gas has decreased. If the temperature of the gas remained constant, then the pressure must have increased according to the Boyle's law;
At constant temperature, the pressure of a given mass of an ideal gas is inversely proportional to its volume.
Salad, a bag of different colored pebbles or sand, etc
C. 28 KJ
AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2
0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
