Answer:
Here's what I get
Explanation:
In a single replacement reaction, one element must displace another element from its compound.
Thus, we must consider the reactivity series, because an element lower down in the series cannot replace one that is above it.
In a double replacement reaction, the cations change partners with the anions. All we need consider are the solubilities of the new cation-anion partners.
Answer:
Warm front
Explanation:
A warm front forms when a warm air mass pushes into a cooler air mass, shown in the image to the right (A). Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms. Warm fronts move more slowly than cold fronts because it is more difficult for the warm air to push the cold, dense air across the Earth's surface. Warm fronts often form on the east side of low-pressure systems where warmer air from the south is pushed north.
You will often see high clouds like cirrus, cirrostratus, and middle clouds like altostratus ahead of a warm front. These clouds form in the warm air that is high above the cool air. As the front passes over an area, the clouds become lower, and rain is likely. There can be thunderstorms around the warm front if the air is unstable.
On weather maps, the surface location of a warm front is represented by a solid red line with red, filled-in semicircles along it, like in the map on the right (B). The semicircles indicate the direction that the front is moving. They are on the side of the line where the front is moving. Notice on the map that temperatures at ground level are cooler in front of the front than behind it.
One mole of a substance contains 6.02×10∧23 particles,
1 mole of a aluminium contains 27 g
35 g of aluminium contains 35/27 =1.296 moles
Thus, the number of particles will be 1.296 × 6.02 ×10∧23
= 7.804 × 10∧23 particles,
Hence, 35 g of Aluminium contains 7.804 × 10∧23 atoms
Answer is: the average atomic mass 217.606 amu.
Ar₁= 203.973 amu; the average atomic mass of isotope.
Ar₂ = 205.9745 amu.
Ar₃ = 206.9745 amu.
Ar₄ = 207.9766 amu.
ω₁ = 1.40% = 0.014; mass percentage of isotope.
ω₂ = 24.10% = 0.241.
ω₃ = 22.10% = 0.221.
ω₄ = 57.40% = 0.574.
Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.
Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.
Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.
Ar = 217.606 amu.
But abundance of isotopes is greater than 100%.
It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.
