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25000 grams.
The ratio (for the covalent radius of the atom) is cca. 1/25000.
Assume, that you digested 500 mg of the soil in acid, and the final volume of your sample is 25 ml. The solution was measured on ICP-AES and 0.5 mg/L Cu was found. How much Cu the soil contains? 25000 micro liter / 500 mg (or 25 ml / 0.5 g) = 50. This is your dilution factor. Multiply your measured result by this factor 0.5 mg/L Cu * 50 = 25 mg Cu per 1 kg of the soil. Of course, if you diluted your digested sample during the measurement, this dilution factor also must be taken into account.
1 1 .There are 6 rules for Significant figures.(SF). Non zero digits count as sf. Ex 34.5=3 Sf and 4367=4 Sf. 2. Leading zeros never count Ex 0.34= 2 Sf and 0.00399 = 3 sf 3. Zeros between sf count ex 405=3sf and .0402=3 sf and 20000.1=6 sf 4.IF a number shows a decimal point ending zeros count. Ex 25.0=3 sf and 10.0= 3 sf and 0.004500=4 sf 5. If a number does not show a decimal point, ending zeros dont count. Ex 500=1 sf and 25000 = 2 sf and 10= 1 sf 6. If a number is written is correct scientific notation, then all digits in fornt of (x10) count. Ex 2.4x10^5 = 2 sf and 3.00x10^8 = 3sf
25 L
The question cannot be answered sensibly. 25000 m3 is a measure of volume, not of mass. That volume of lead, for example, will have quite a different mass to the same volume of air. In addition to the volume, you need to know the density of the substance involved.
That is 25 litres.
To calculate the density, you need to divide the mass by the volume.
250003 = 25000 x 25000 x 25000 = 15625000000000
15% of 25,000= 15% * 25000= 0.15 * 25000= 3,750
25000*25000=175000000
15% of 25000= 15% * 25000= 0.15 * 25000= 3750
Percent or percentage is equivalent to hundredths. 8% of 25000 is 25000 x 8/100 = 2000. 25000 + 8% of 25000 = 25000 + 2000 = 27000 NOTE : As 25000 is 100% then the problem can be solved by finding 108% of 25000. 25000 x 108/100 = 27000.
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