Answer:
A) positive; added
Explanation:
Based on the reaction:
2NaHCO3(s) + 129kJ → Na2CO3(s) + H2(g) + CO2(g)
<em>2 moles of NaHCO3 requires 129kJ to produce 1 mole of Na2CO3, 1 mole of H2 and 1 mole of CO2.</em>
<em />
That means, the energy must be added being, thus, an exothermic reaction. The exothermic reactions have ΔH >0.
Thus, right answer is:
A) positive; added
Answer:
Conversion factor;
Molar mass;
Avogadro's constant and molar mass
Explanation:
- Firstly, an intermediate step is to define the conversion factor that will be then used in a conversion technique called dimensional analysis in order to convert from one unit to another. An example of a conversion factor would be, for example, 1 L = 1000 mL, which can be manipulated as a fraction, either
or 
; - Secondly, in order to convert mass to moles, we need to know the molar mass of a compound which has a units of g/mol (that is, it shows how many grams we have per 1 mole of substance.
- Thirdly, Avogadro's constant,
tells us that there is 
number of molecules or atoms in 1 mole of substance. We need two conversion factors to convert the number of molecules to a mass: firstly, we need to convert the number of molecules into the number of moles using Avogadro's constant and then we need to use the molar mass to convert the moles obtained into mass.
<span>1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol
The independent variable would be ethylene glycol and dependent variable would be density.
A. A 25-mL volumetric flask with its stopper has a mass of 32.6341 g. The same flask filled to the line with ethylene glycol (C2H6O2, automotive antifreeze) solution has a mass of 58.0091 g. What is the density of the ethylene glycol solution?
Density = 58.0091 - 32.6341 / .025 = 1015 g/L
B. What is the molarity of the ethylene glycol solution, if the mass of ethylene glycol in the solution is 12.0439 g?
Molarity = 12.0439 ( 1 mol / 62.07 g) / 0.025 = 7.8 M</span>
Missing question: A possible atomic weight of Q is?
A)12.7.
B)19.0.
C)27.5
D)38.0
E)57.0.
Answer is: possible atomic weight of Q is B) 19.0.
Molecular weight of first molecule is 38.0 g/mol.
Molecular weight of second molecule is 57.0 g/mol.
Molecular weight of third molecule is 76.0 g/mol.
Molecular weight of fourth molecule is 114.0 g/mol.
All four molecular weights can be division only by 19 to get whole number:
Molecule 1 (38 ÷ 19 = 2), molecule 2 (57 ÷ 19 = 3), molecule 3 (76 ÷ 19 = 4) and molecule 4 ( 114 ÷ 19 = 6).
Molecular formulas are than Q₂, Q₃, Q₄ and Q₆.
