<span>1.15x10^24 molecules of hypothetical substance b
Making the assumption that each molecule in hypothetical substance a reacts to produce a single molecule of hypothetical substance b, then the number of molecules of substance b will be the number of moles of substance a multiplied by avogadro's number. So
Moles hypothetical substance a = 29.9 g / 15.7 g/mol = 1.904458599 moles
This means that we should also have 1.904458599 moles of hypothetical substance b. And to get the number of atoms, multiply by 6.0221409x10^23, so:
1.904458599 * 6.0221409x10^23 = 1.146892x10^24 molecules.
Rounding to 3 significant figures gives 1.15x10^24</span>
Answer:
Explanation:
The standard system of measurement is what is popularly referred to as SI unit (or international system of units). For example, the SI unit for length is basically in meters (m), which is convertible to millimeter (mm), centimeter (cm) or even kilometer (km).
If there is no standard system of measurement,
1) Scientists will not be able to compare data as every scientist will use a unit that suits him/herself which might make data incomparable as some units might be difficult to convert to other units or there conversion methods might even be debatable.
2) There might be a general lack of understanding of the research work done by a scientist/scientists in the scientific community. This is because the standard system of measurement affords scientists to communicate properly with the same or convertible units and based on 1. above, if scientists use any unit that isn't agreed upon, different/unknown/debatable units might make understanding of research works incomprehensible.
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c

0.511
Now using the equilibrium constant 
![K_c = \dfrac{[F6P]}{[G6P]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5BG6P%5D%7D)
![0.511 = \dfrac{[F6P]}{[2.65]}](https://tex.z-dn.net/?f=0.511%20%3D%20%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5B2.65%5D%7D)
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
Answer:
B.FeCl₂
Explanation:
there is twice as much chlorine as there is iron