The name of the compound Na2SO4 is:
nitrogen sulfite
nitrogen sulfide
sodium sulfate
sodium sulfite
nickel sulfide
sodium sulfate
The formula of calcium phosphate is:
CaPO4
CaP
Ca3(PO4)2
Ca4(PO4)3
Ca(PO4)2
Ca3(PO4)2
The formula for calcium acetate is:
Ca2C2H3O2
CaC2O4
Ca(C2O4)2
Ca(C2H3O2)2
CaC2H3O2
Ca(C2H3O2)2
Answer:
c) Seasonal migration
Explanation:
Because A, B and C are all examples of structural adaptations. And seasonal migration is not something that is in an animals structure but is instinct.
Have a great day!
<span>Answer is: 2940 mL of
the HCL solution.</span>
c₁(HCl) = 10.0 M.
V₂(AgNO₃<span>) = ?.
c</span>₂(AgNO₃<span>) = 0.85 M.
V</span>₁(AgNO₃<span>) = 250 mL </span>÷ 1000 mL/L = 0.25 L.
<span>
c</span>₁<span> - original concentration of the solution, before it
gets diluted.
c</span>₂<span> - final concentration of the solution, after dilution.
V</span>₁<span> - volume to be diluted.
V</span>₂<span> - final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
V</span>₂(HCl) = c₁ · V₁ ÷ c₂.
<span>
V</span>₂(HCl) = 10 M · 0.25 L ÷ 0.85 M.
<span>
V</span>₂(HCl) = 2.94 L ·
1000 mL = 2940 mL.
2.4368 MILOGRAMS is burned
Answer:
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).
Explanation:
The reaction between an acid and a base is called neutralization, forming a salt and water.
Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.
When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:
V acid *M acid = V base *M base
where V represents the volume of solution and M the molar concentration of said solution.
In this case:
- V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
- M acid= 0.129 M
- V base= ?
- M base= 0.135 M
Replacing:
0.0137 L* 0.129 M= V base* 0.135 M
Solving:

V base=0.0131 L = 13.1 mL
<u><em>
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>