Answer: 3.0 kJ × 1 mol/40.65 kJ× 18.02 g/mol × 1 mL/1 g= 1.3 mL
<span>Atomic number of magnesium is 12. That implies that in ground state magnesium has 2 electrons in the first level, 8 electrons in the second level, and 2 electrons in the third level. That is represented by 2 8 2. That the atom is in an excited state means that one electron (at least) is a upper level than where it is in grounded state. That situation is represented by the option (1) 2 7 3, where one electron from the second level has been promoted to the third level.</span>
Answer:
12
Explanation:
alkyne formula is CnH2n-2
heptyne = 7
Answer:
H₂O + CO₂ → H₂CO₃
Option D is correct.
Law of conservation of mass:
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two H and two O atoms present on left side while on right side only one O and two H atoms are present so mass in not conserved. This option is incorrect.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side of equation while on right side two H, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. This option is correct.
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
