Answer:
A Winter storm usually does not come with a warm fronthey warm front.
Explanation:
When the front passes over an area it means to change in the weather. Many. Many friends cause weather events such as rain thunderstorms gusty winds tornadoes and hurricanes.
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
First, find how many grams are in 1 mole of water.
For a hydrogen atom, there is about 1 gram per mole. For an oxygen atom, there are about 16 grams per mole.
In H2O, there are two hydrogen atoms and one oxygen atom. This means there are 18 grams in one mole of water. Multiply the mass in one mole by your number of moles.
18 x 11.8 = 212.4 grams
You have 212.4 grams of water.
Answer:
The answer to your question is: ΔH = -283 kJ/mol, first option
Explanation:
Reaction
CO + O₂ ⇒ CO₂
ΔH = ∑H products - ∑H products
ΔH = -393.5 - (-110.5 + 0)
ΔH = -393.5 + 110.5
ΔH = -283 kJ/mol
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.
