Answer:
The answer to your question is below
Explanation:
A.
[H₃O⁺] = 2 x 10⁻¹⁴ M
pH = ?
Formula
pH = - log [H₃O⁺]
Substitution
pH = - log [2 x 10⁻¹⁴]
Result
pH = 13.7
B.
[H₃O⁺] = ?
pH = 3.12
Formula
pH = - log [H₃O⁺]
Substitution
3.12 = - log [H₃O⁺]
![10^{-3.12} = [H_{3} O^{+}]](https://tex.z-dn.net/?f=10%5E%7B-3.12%7D%20%3D%20%5BH_%7B3%7D%20O%5E%7B%2B%7D%5D)
Result
[H₃O⁺] = 7.59 M
Mass Molar of
Ca = 3*40 = 120 amu
P = 2*31= 62 amu
O = (16*4)*2 = 64*2 = 128 amu
--------------------------------------
Mass Molar of
= 120 + 62 + 128 = 310 g/mol
Therefore: <span>What is the gram formula mass of Ca3(PO4)2 ?
</span>Answer:
310 grams
Your answer is 3.25 moles of Bromine
Answer: B- Chemical bonds are formed. Energy is released in the form of heat.
Explanation: I hoped that helped !
Answer:
Option C
CH₃CH₂CH₂COOH
Explanation:
Carbonxylic acids are compounds which has the general formula
R–COOH where R is an alkyl group.
Considering the options given in the question above,
For A:
CH₃CH₂OCH₂CH₃ is an ether compound with general formula ROR' where R and R' are both alkyl group.
For B:
CH₃CH₂CH₂CH₂OH is an alcohol with general formula ROH where R is an alkyl group.
For C:
CH₃CH₂CH₂COOH is a carbonxylic acid with general formula R–COOH where R is an alkyl group.
For D:
CH₃CH₂C=OCH₂CH₃ is a ketone compound with general formula RC=OR' where R and R' are both alkyl group
For E:
ClCH₂CH₂CH₂CH₂CH₂CH₂Br is simply an Alkyl halide with general formula XRX where X is an halogen (i.e F, Cl, Br or I) and R is an alkyl group.
From the above illustration, only option C contains a Carbonxylic compound.
