<h3>
Answer:</h3>
4 cm³
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
<u>Chemistry</u>
<u>Gas Laws</u>
Density = Mass over Volume
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
D = 25 g/cm³
m = 100 g
<u>Step 2: Solve for </u><em><u>V</u></em>
- Substitute variables [D]:
- Multiply <em>V</em> on both sides:
- Isolate <em>V</em>:
Answer:
3.955*10^48
Explanation:
1 mole of a substance gives 6.02*10^23/6.57*10^24 will give x then cross multiply the answer. is 3.955*10^48
Answer:
The answer to your question is 2 moles
Explanation:
Data
mass of H₂O = 36 g
moles of H₂O = ?
Process
1.- Calculate the molar mass of water (H₂O)
H₂O = (1 x 2) + (16 x 1) = 2 + 16 = 18 g
2.- Use proportions and cross multiplication to find the answer.
18 g of H₂O ---------------- 1 mol
36 g of H₂O --------------- x
x = (36 x 1) / 18
x = 36/18
x = 2 moles
A low pressure system has lower pressure at its center than the areas around it. Winds blow towards the low pressure, and the air rises in the atmosphere where they meet. As the air rises, the water vapor within it condenses, forming clouds and often precipitation.
<u>Explanation</u>:
- Wind flow towards the low pressure and the air rises in the atmosphere. As the air increases, the water vapor within it solidifies, forming clouds and undergo precipitation. Low pressure formed in the center areas.
- The atmospheric circulations of air up and down in a low-pressure area remove a small amount of atmosphere. This usually happens between warm and cold air masses by flowing air which tries to reduce the contrast of temperature.
