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n200080 [17]
2 years ago
10

If the titration of h2po4− in a urine sample was continued until all of the acid in the solution was neutralized, how many equiv

alents of naoh would be needed to fully neutralize the solution?
Chemistry
1 answer:
sergiy2304 [10]2 years ago
5 0

Two equivalents of NaOH are needed to fully neutralize the solution.

<h3>What is a neutralization reaction?</h3>

A neutralization reaction involves the reaction of an acid and a base to form salt and water. The reaction is exothermic in nature.

For example,

HCl (aq) + NaOH(aq)\rightarrow NaCl(aq) + H_2O (l)

During the titration of H_2PO_4^- it gives 2H^+.

H_2PO_4^- \rightarrow 2H^+ +PO_4^3^-

To neutralize these 2H^+, 2 OH^- is needed which come from 2 equivalents of NaOH

2NaOH \rightarrow2 Na^+ +2OH^-

Therefore, 2 equivalents of NaOH are needed

Learn more about neutralization reaction:

brainly.com/question/23008798

#SPJ4

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Give 3 example<br>of a<br>solid that sublimes.​
tino4ka555 [31]

Here are a few examples :)

iodine (I2)

naphthalene

aresenic (As)

ferrocene

water (H2O)

carbon dioxide (CO2)

Hope this helps :)

5 0
3 years ago
How many molecules are present in 2 H₂ SO4​
fomenos

Answer:

in H2So4 splits in 2 H+ and a

so42 -particle=3particle per mole.So2

moles H2So4 will result in 3*2=6moles

of molecules.

5 0
3 years ago
Read 2 more answers
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
Vitamin C contains only carbon, hydrogen, and oxygen. When a 1.00 g was combusted, 1.4991 g of CO2 and 0.4092 g of H2O were obta
Alina [70]

The empirical formula for this vitamin : C₃H₄O₃

<h3>Further explanation   </h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

  • Determine the mass ratio of the constituent elements of the compound.  
  • Determine the mole ratio by dividing the percentage by the atomic mass

Mass of C in CO₂ :(MW C = 12 g/mol, CO₂=44 g/mol)

\tt \dfrac{12}{44}\times 1.4991=0.409~g

Mass of H in H₂O :(MW H = 1 g/mol, H₂O = 18 g/mol)

\tt \dfrac{2.1}{18}\times 0.4092=0.0455~g

Mass O = Mass sample - (mass C + mass H) :

\tt 1-(0.409+0.0455)=0.5455~g

mol ratio C : H : O =

\tt \dfrac{0.409}{12}\div \dfrac{0.0455}{1}\div \dfrac{0.5455}{16}\\\\0.0341\div 0.0455\div 0.0341\rightarrow 1\div 1.33\div 1=3\div 4\div 3

5 0
3 years ago
Please help me<br> 5. Where is potential energy decreasing?<br> А B C D
Nataly [62]

Answer:

D

Explanation:

I believe it is D. your kinetic energy would be at b. A, the cart would be going at a constant rate, because there is no hill or steep slope.

7 0
2 years ago
Read 2 more answers
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