Lithium has 1 valence electron available for bonding. So its A.
Answer:
1. Potassium, K.
2. Calcium, Ca.
3. Gallium, Ga.
4. Carbon, C.
5. Bromine, Br.
6. Barium, Ba.
7. Silicon, Si.
8. Gold, Au.
Explanation:
Atomic radius can be defined as a measure of the size (distance) of the atom of a chemical element such as hydrogen, oxygen, carbon, nitrogen etc, typically from the nucleus to the valence electrons. The atomic radius of a chemical element decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon). Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).
Additionally, the unit of measurement of the atomic radius of chemical elements is picometers (1 pm = 10 - 12 m).
1. Li or K: the atomic radius of lithium is 167 pm while that of potassium is 243 pm.
2. Ca or Ni: the atomic radius of calcium is 194 pm while that of nickel is 149 pm.
3. Ga or B: the atomic radius of gallium is 136 pm while that of boron is 87 pm.
4. O or C: the atomic radius of oxygen is 48 pm while that of carbon is 67 pm.
5. Cl or Br: the atomic radius of chlorine is 79 pm while that of bromine is 94 pm.
6. Be or Ba: the atomic radius of berryllium is 112 pm while that of barium is 253 pm.
7. Si or S: the atomic radius of silicon is 111 pm while that of sulphur is 88 pm.
8. Fe or Au: the atomic radius of iron is 156 pm while that of gold is 174 pm.
Answer:
The correct answer is option B.
Explanation:
Michaelis–Menten 's equation:
![v=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D%3Dk_%7Bcat%7D%5BE_o%5D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
![V_{max}=k_{cat}[E_o]](https://tex.z-dn.net/?f=V_%7Bmax%7D%3Dk_%7Bcat%7D%5BE_o%5D)
v = rate of formation of products
[S] = Concatenation of substrate = ?
= Michaelis constant
= Maximum rate achieved
= Catalytic rate of the system
= initial concentration of enzyme
We have :
![v=\frac{V_{max}}{4}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7BV_%7Bmax%7D%7D%7B4%7D)
[S] =?
![K_m=0.0050 M](https://tex.z-dn.net/?f=K_m%3D0.0050%20M)
![v=V_{max}\times \frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
![\frac{V_{max}}{4}=V_{max}\times \frac{[S]}{(0.0050 M+[S])}](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7Bmax%7D%7D%7B4%7D%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%280.0050%20M%2B%5BS%5D%29%7D)
![[S]=\frac{0.005 M}{3}=1.7\times 10^{-3} M](https://tex.z-dn.net/?f=%5BS%5D%3D%5Cfrac%7B0.005%20M%7D%7B3%7D%3D1.7%5Ctimes%2010%5E%7B-3%7D%20M)
So, the correct answer is option B.
Answer:
i.e. mass of 1 mole of glucose, C6H12O6 = (6 × 12.01 + 12 × 1.01 + 6 × 16.00) g = 180.18 g (using atomic weight data to 2 decimals) 1 mole of carbon atoms weighs 12.01 g and there are 6 moles of C atoms in 1 mole of glucose, so the mass of carbon in 1 mole of glucose = 6 × 12.01 g = 72.06 g.
Answer:
The solvent
Explanation:
After the solid is recoverd by decanting the liquid, some solvent particles may still adhere to the solid.
In order to remove the solvent that may have adhered to the solid completely , the solid is washed before it is now dried.