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.
“It can hold two electrons”
Answer:
Biodiversity: Bio meaning life and diversity meaning variety. This word refers to the different variations of life that can be found on Earth (plants, animals, micro-organisms, fungi).
Explanation:
1. The mass of 1.33×10²² mole of Sb is 1.62×10²⁴ g
2. The mass of 4.75×10¹⁴ mole of Pt is 9.26×10¹⁶ g
3. The mass of 1.22×10²³ mole of Ag is 1.32×10²⁵ g
4. The mass of 9.85×10²⁴ mole of Cr is 5.12×10²⁶ g
<h3>1. Determination of the mass of 1.33×10²² mole of Sb</h3>
- Mole of Sb = 1.33×10²² mole
- Molar mass of Sb = 122 g/mol
Mass = mole × molar mass
Mass of Sb = 1.33×10²² × 122
Mass of Sb = 1.62×10²⁴ g
<h3>2. Determination of the mass of 4.75×10¹⁴ mole of Pt</h3>
- Mole of Pt = 4.75×10¹⁴ mole
- Molar mass of Pt = 122 g/mol
Mass = mole × molar mass
Mass of Pt = 4.75×10¹⁴ × 195
Mass of Pt = 9.26×10¹⁶ g
<h3>3. Determination of the mass of 1.22×10²³ mole of Ag</h3>
- Mole of Ag = 1.22×10²³ mole
- Molar mass of Ag = 108 g/mol
Mass = mole × molar mass
Mass of Ag = 1.22×10²³ × 108
Mass of Ag = 1.32×10²⁵ g
<h3>4. Determination of the mass of 9.85×10²⁴ mole of Cr</h3>
- Mole of Cr = 9.85×10²⁴ mole
- Molar mass of Cr = 52 g/mol
Mass = mole × molar mass
Mass of Cr = 9.85×10²⁴ × 52
Mass of Cr = 5.12×10²⁶ g
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