When sodium carbonate is dissolved in water, it doesn’t dissociate, or break up, completely. There is always some sodium carbona
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Hello!
We have the following data:
Area (A) = 50 square feet
Mass (m) = 8.5 ounces
Density (d) = 2.70 g/cm³
Volume (V) = ?
Thickness (T) =? (in mm)
To move on, we must transform the area of 50 ft² in cm², let's see:
1 ft² ------- 929,0304 cm²
50 ft² ----- A
In the same way, we will convert the mass of 8.5 oz in grams, see:
1 oz -------- 28,3495 g
8,5 oz ------- m
Knowing that the density is 2.70 g/cm³ and the mass is 240.97075 g, we will find the volume, applying the data in the density formula we have:
The statement wants to find the thickness of the packaging, for this we have some important data, such as: V (volume) = 89,25 cm³ and Area (A) = 46451,52 cm² and T (thickness) =? (in mm)
In the calculations of Costs in Surface Treatment of a part within the flat geometry, we will use the following formula:
We will convert to millimeters, going through a decimal place on the right
Hope this helps! :))
We have the following data:
Area (A) = 50 square feet
Mass (m) = 8.5 ounces
Density (d) = 2.70 g/cm³
Volume (V) = ?
Thickness (T) =? (in mm)
To move on, we must transform the area of 50 ft² in cm², let's see:
1 ft² ------- 929,0304 cm²
50 ft² ----- A
In the same way, we will convert the mass of 8.5 oz in grams, see:
1 oz -------- 28,3495 g
8,5 oz ------- m
Knowing that the density is 2.70 g/cm³ and the mass is 240.97075 g, we will find the volume, applying the data in the density formula we have:
The statement wants to find the thickness of the packaging, for this we have some important data, such as: V (volume) = 89,25 cm³ and Area (A) = 46451,52 cm² and T (thickness) =? (in mm)
In the calculations of Costs in Surface Treatment of a part within the flat geometry, we will use the following formula:
We will convert to millimeters, going through a decimal place on the right
Hope this helps! :))