Answer:
- <em>The maximum amount of copper allowed in 100 g of water is </em><u><em>0.00013 g</em></u>
Explanation:
To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg) and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):
First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.
Now, set the proportion:
- 1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O
Solve for x:
- x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu
Convert mg to grams:
- 0.13 mg × 1 g / 1,000 mg = 0.00013 g
Answer: 0.00013 g of copper.
Answer:
There was an improvement in accuracy. There was no change in precision.
Explanation:
<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).
The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.
Too much money and dangerous
Kp= (COCl2)/[(CO)(Cl2)]= 1.49 x 10^8
1.49 x 10^8= (COCl2/((2.22x10-4)(2.22x10-4))
COCl2= 1.49x10^8 x ((2.22x10-4)(2.22x10-4))= 7.34 atm
Step one: Identify reactants and products and place them in a word equation.
Step two: Convert the chemical names into chemical formulas. Place them based on the chemical equation and write the state symbols.
Step three: balance the chemical equation.
