Answer:
0.144M
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
HNO3 + KOH —> KNO3 + H20
From the equation,
nA = 1
nB = 1
From the question given, we obtained the following:
Ma =?
Va = 30.00mL
Mb = 0.1000M
Vb = 43.13 mL
MaVa / MbVb = nA/nB
Ma x 30 / 0.1 x 43.13 = 1
Cross multiply to express in linear form
Ma x 30 = 0.1 x 43.13
Divide both side by 30
Ma = (0.1 x 43.13) /30 = 0.144M
The molarity of the nitric acid is 0.144M
Answer:
no
Explanation:
Radium is silvery, lustrous, soft, intensely radioactive. It readily oxidizes on exposure to air, turning from almost pure white to black. Radium is luminescent, corrodes in water to form radium hydroxide. Although is the heaviest member of the alkaline-earth group it is the most volatile.
Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
Answer:
Option C = same period.
Explanation:
All these elements belongs to second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.
Electronic configuration of lithium:
Li₃ = [He] 2s¹
Electronic configuration of beryllium:
Be₄ = [He] 2s²
Electronic configuration of boron:
B₅ = [He] 2s² 2p¹
Electronic configuration of carbon:
C₆ = [He] 2s² 2p²
Electronic configuration of nitrogen:
N₇ = [He] 2s² 2p³
Electronic configuration of oxygen:
O₈ = [He] 2s² 2p⁴
Electronic configuration of fluorine:
F₉ = [He] 2s² 2p⁵
Electronic configuration of neon:
Ne₁₀ = [He] 2s² 2p⁶
All these elements present in same period having same electronic shell.
However their families, valance electrons and group are different. Boron have three valance electrons and belongs to group 3A. Carbon belongs to group 4A and have 4 valance electrons. Nitrogen belongs to group 5A and have five valance electrons. Oxygen belongs to group 6A and have six valance electrons. Fluorine belongs to group 7A and have seven valance electrons.
Natural gas is primarily composed of methane (CH4)
Natural gas is a naturally occurring hydrocarbon mixture which is primarily composed of Methane(CH4), but it also contains ethane,propane and heavier hydrocarbon. In addition it contain small amount of nitrogen, carbon dioxide,hydrogen sulfide and traces amount of water.
