Answer:
A producer is a plant or fungi but it mainly is a organism that produces food for a primary consumer
Explanation:
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
Answer:
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced
Now use the ideal gas law to find the volume V.
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm
T = 21 C + 273 = 294 K
PV = nRT
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L
<u>Answer:</u> The balanced half reaction is written below.
<u>Explanation:</u>
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
When oxygen gas is reduced to oxide ions, the number of electron transferred are 2
The chemical equation for the reduction of oxygen gas to oxide ions follows:
Hence, the balanced half reaction is written above.
