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 answer to your question is: 16.7 g of KBr
Explanation:
Data
mass KBr = ? g
Volume = 0.400 L
Concentration = 0.350 M
Formula
Molarity = moles / volume
moles = molarity x volume
Process
moles = (0.350)(0.400)
= 0.14
MW KBr = 39 + 80 = 119 g
119 g of KBr -------------------- 1 mol
x -------------------- 0.14 mol
x = (0.14 x 119) / 1
x = 16.7 g of KBr
Answer : The balanced equations will be:
(a) 
(b) 
(c) 
Explanation :
The general rate of reaction is,
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be :
![\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20A%7D%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20B%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20C%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20D%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
![Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
Now we have to determine the balanced equations corresponding to the following rate expressions.
(a) ![Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BCH_4%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BCO_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7Bd%5BCH_3OH%5D%7D%7Bdt%7D)
The balanced equations will be:
(b) ![Rate=-\frac{1}{2}\frac{d[N_2O_5]}{dt}=+\frac{1}{2}\frac{d[N_2]}{dt}=+\frac{1}{5}\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BN_2O_5%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BN_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B5%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The balanced equations will be:
(c) ![Rate=-\frac{1}{2}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[CO_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2CO_3]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BCO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BH_2CO_3%5D%7D%7Bdt%7D)
The balanced equations will be:
Answer:
6
Explanation:
bromine has 1, hydrogen has 4 and nitrogen has 1 therefore equaling 6
