Answer:
Explanation:
This question appear incomplete because of the absence of options. However, places where silver recovery would happen in the United States are states where huge mining of copper, lead, zinc or gold are been mined. Examples of such places include Arizona, California and Oregon; where copper (and some other metals in some cases like gold in Oregon) is heavily mined and silver is recovered as a byproduct.
The precision of a tool can affect the measurement's accuracy. how specific the measurement can get (if it goes to the tenths place, hundredths, etc) can also affect the accuracy.
Answer:
216 g of NO
Explanation:
We begin from the reaction:
4NH₃ + 5O₂ → 4NO + 6H₂O
We determine the limiting reactant with the moles of each reactant:
4 moles of ammonia react to 5 moles of oxygen
Our 7.2 moles of ammonia may react to (7.2 . 5) /4 = 9 moles
It's ok because we have 9.6 moles of oxygen. 0.6 moles still remain.
5 moles of oxygen react to 4 moles of NH₃
Our 9.6 moles of oxygen may react to (9.6 . 4) /5 = 7.68 moles
We only have 7.2 moles of NH₃ and we need 7.68; so there is no enough ammonia and that's our limiting reagent.
Now we determine the moles of product.
4 moles of ammonia can produce 4 moles of NO
Definetely our 7.2 moles, will produce 7.2 moles of oxide.
We convert to mass: 7.2 mol . 30 g/mol = 216 g
<span>Aspirin is the prototypical analgesic used in the treatment of mild to moderate pain. It has anti-inflammatory and antipyretic properties and acts as an inhibitor of cyclooxygenase which results in the inhibition of the biosynthesis of prostaglandins. Aspirin also inhibits platelet aggregation and is used in the prevention of arterial and venous thrombosis. (From Martindale, The Extra Pharmacopoeia, 30th ed, p5)</span>
T<span>his is a straightforward question related to the surface energy of the droplet. </span>
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>