MA= output force/input force
MA= 100N/20N
MA= 50
If iron has a density of 7.87g/cm³ and a mass of 3.729g, then the volume of iron is 0.474cm³
HOW TO CALCULATE VOLUME:
- The volume of a substance can be calculated by dividing the mass by its density. That is;
Volume (mL) = mass (g) ÷ density (g/mL)
- The density of iron is given as 7.87g/cm³ while its mass is 3.729g of iron. Hence, the volume can be calculated as follows:
Volume = 3.729 ÷ 7.87
Volume = 0.474cm³
Therefore, the volume of iron is 0.474cm³
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Insulation wraps because independent is the variable you are changing to affect the dependent variable (what you are measuring)
Answer:
5 electrons
Explanation:
When an atom loses are gain the electrons ions are formed.
There are two types of ions.
Anion
Cation
1 = Anion
It is formed when an atom gain the electrons. when atom gain electron negative charge is created on atom. For example.
X + e⁻ → X⁻
2= Cation
It is formed when an atom loses the electrons. when atom lose electron positive charge is created on atom. For example.
X → X⁺ + e⁻
When an atom loses three electrons +3 charge is created. If this atom have 8 protons it means there were 8 electrons too. When it lost three electrons then remaining number of electrons are 5.
X → X³⁺ + 3e⁻
The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
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