There are an infinity of possible solutions. Here are some: 1 cm * 1 cm * 550 cm 10 cm * 10 cm * 5.5 cm 1 m * 1 m * 0.055 cm or 1 cm * 2 cm * 275 cm 1 cm * 20 cm * 27.5 cm
110 cm + 1 cm = 111 cm
1 m = 100 cm 1 cm = 1 % from 1m 1 cm = 1 m/100
Its area is: 1 times 1 = 1 square cm
Multiplication of length means expanding in dimension (whatever that means) that is 1 cm x 1 cm = 1 cm^2 it's 2 dimention, so it's area of a 1 by one rectangle. 1 cm x 1 cm x 1 cm = 1 cm ^3 which is a cube, it's 3D DO NOT try to visualize the following 1 cm x 1 cm x 1 cm x 1 cm = 1 cm^4, it's a 4D model (no one knows what it looks like) so in your case, 2 cm x 3 cm x 2 cm= (2 x 3 x 2) cm ^3 = 12 cm ^3 (paper form is cm to the power of 3)
To solve this problem, we basically have 2 equations and 2 unknowns. The unknowns are the (volume of water) & the (volume of 70 wt%) nitric acid to add. * This problem will assume that you are interested in making 1 L (or 1000 mL) of 5 wt% nitric acid solution. Equation 1: (volume of water) + (volume of 70 wt% nitric acid) = 1000 mL Equation 2: mass of nitric acid / [mass of water + mass of 70 wt% nitric acid solution] = 0.05 (0.05 is 5 wt%) * Remember that mass = density * volume * Remember that 70 wt% nitric acid solution mean that for 100 grams (gm) of this acid, then there's 70 grams of HNO3 * Remember that density of 70 wt% nitric acid solution is 1.413 gm/cm^3 * Remember that density of water is 1 gm/cm^3 Equation 2 is now re-written as: [(density of 70 wt% nitric acid soln)*(volume of 70 wt% nitric acid)*0.70] / [(volume water)*(1gm*cm^3) + (volume of 70 wt% nitric acid)*(1.413gm/cm^3)] = 0.05 Solving for the 2 equations gives answer to the 2 unknowns: Answer: To make 1000 mL of 5 wt% nitric acid solution, add 1) 51.63 mL of 70 wt% nitric acid solution 2) 948.37 mL of water
It depends on the kind of fruit, usually. But on average, 1/2 of a cup is about a serving.
w2 - w3 / w2 -w1 x 100 w1 - wt of empty bottle w2 - wt of bottle with sample w3 - wt of bottle with sample after drying.
2.2046 pounds = 1 kg.
according to the KNM-WT 15000 (Nariokotome) skeleton, discovered in 1984. This skeleton was estimated to have a moderately tall adult stature -- around 185 cm (6 feet 1 inch).
There are an infinity of possible solutions. Here are some: 1 cm * 1 cm * 550 cm 10 cm * 10 cm * 5.5 cm 1 m * 1 m * 0.055 cm or 1 cm * 2 cm * 275 cm 1 cm * 20 cm * 27.5 cm
110 cm + 1 cm = 111 cm
1 foot = 30.48 cm.1 foot = 30.48 cm.1 foot = 30.48 cm.1 foot = 30.48 cm.
1 m = 100 cm 1 cm = 1 % from 1m 1 cm = 1 m/100
16 wt oz = 1 Lb
10 cm as a fraction of 1 meter = 1/10 1 meter = 100 cm 10 cm/1 meter = 10 cm/100 cm = 1/10
1 m = 100 cm 1 m² = 1 m × 1 m = 100 cm × 100 cm = 10,000 cm²