A word equation is a chemical reaction described using words.
A common example is the act of photosynthesis - the process plants use to make glucose (sugar) to use as 'food'.
Plants convert water and carbon dioxide into oxygen and glucose.
A word equation to express this is:
Water + Carbon Dioxide → Glucose + Oxygen
The other type of equation is a symbol equation - this uses the symbols of the elements instead of the common names:
H₂O + CO₂ → C₆H₁₂O₆ + O₂
There is also a balanced version:
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
<em>If you want information on the balanced symbol equations, feel free to PM me.</em>
Answer:
The lung
Explanation:
The model of the respiratory system made by Megan consists of two balloons. The first balloon stretched across the bottom of the bottle represents the diaphragm which contracts and relaxes to allow air in and out of the lungs. The balloon inside the bottle represents one lung.
Breathing in causes the balloon inside the bottle to be filled with air. This is preceded by the expansion of the diaphragm which makes the lungs to be filled with air. Breathing out causes a contraction of the diaphragm thus making the lungs to let out air.
Explanation:
It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.
or 
So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.
Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
