So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
that 's what I think
Answer:
B. use a new pair of gloves for each piece of evidence
Explanation:
When it comes to the basic precautions of handling biological evidence samples, one has to prevent cross contamination. This refers to the<em> transfer of DNA from one evidence to another evidence</em>; thus, it is important for Jane to use a new pair of gloves for each piece of evidence in order to prevent such occurrence.
Wearing of gloves will also <em>prevent Jane's DNA from being implanted into the sample</em>, and it will keep her safe from contracting blood-borne pathogens like those in the saliva, blood or semen.
Answer:
3.0 moles Al₂O₃
Explanation:
We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.
4 Al + 3 O₂ -----> 2 Al₂O₃
6.0 moles Al 2 moles Al₂O₃
---------------------- x ------------------------- = 3.0 moles Al₂O₃
4 moles Al
4.0 moles O₂ 2 moles Al₂O₃
---------------------- x ------------------------- = 2.7 moles Al₂O₃
3 moles O₂
As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.
However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.
Answer:
Mass percentage → 0.074 %
[F⁻] = 741 ppm
Explanation:
Aqueous solution of flouride → [F⁻] = 0.0390 M
It means that in 1L of solution, we have 0.0390 moles of F⁻
We need the mass of solution and the mass of 0.0390 moles of F⁻
Mass of solution can be determined by density:
1g/mL = Mass of solution / 1000 mL
Note: 1L = 1000mL
Mass of solution: 1000 g
Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g
Mass percentage → (Mass of solute / Mass of solution) . 100
(0.741 g / 1000 g) . 100 = 0.074 %
Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)
0.741 g . 1000 mg/1g = 741 mg
1000 g . 1 kg/1000 g = 1kg
741 mg/1kg = 741 ppm
