The balanced equation for
Ca(OH)2 + H3PO4→ Ca3(PO4)2 + H2O is
3 Ca(OH)2 +2 H3PO4→ Ca3(Po4)2 + 6 H2O
3 moles of Ca(OH)2 reacted with 2 moles of H3PO4 to form 1 mole of Ca3(PO4)2 and 6 moles of H2O
Answer:
<em>Your</em><em> </em><em>Answer</em><em> </em><em>is</em><em> </em><em>Option</em><em> </em><em>A</em><em> </em><em>that</em><em> </em><em>is</em><em> </em><em>Gas</em><em>.</em>
Answer:
Formic acid, citric acid, Oxalic acid, washing soda, baking soda, etc. can be some examples of natural acids and natural bases. They both have domestic, industrial, and various other purposes.
Explanation:
<h3><u>
NATURAL ACIDS</u>
:</h3>
There are lots of natural acids present in our nature. Some of them are the following:
> <u>Formic acid</u>
USE: It is used in the stimulation of oil and gas wells as it is less reactive towards the metal.
> <u>Citric acid</u>
USE: It is considered as the best rust remover as it doesn't harm the metal just remove the rust.
> <u>Oxalic acid</u>
USE: It easily remove iron and ink stains and that's why it is used as an acid rinsing material in Laundries.
<h3><u>
NATURAL BASES</u>
:</h3>
There is a variety of natural base found in our nature which founds a lot of uses in day to day life. some of them are the following:
> <u>Washing soda</u>
USE: It is used in commercial detergent mixture to treat hard water.
> <u>Baking soda</u>
USE: It is the best rising agent used mostly in cooking and for domestic purposes like removing stains, etc..
The answer should be Pure substance
Answer:
V₂ = 1.5 L
Explanation:
Given data:
Initial volume of balloon = 1.76 L
Initial temperature = 295 K
Final temperature = 253.15 K
Final volume = ?
Solution:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 1.76 L ×253.15 K / 295 K
V₂ = 445.54 L.K /295 K
V₂ = 1.5 L
