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Vesnalui [34]
2 years ago
5

What volume of 0.120 M HNO3(aq) is needed to completely neutralize 150.0 milliliters of 0.100 M NaOH(aq)?

Chemistry
1 answer:
True [87]2 years ago
7 0
HNO3 is a strong acid and NaOH is a
strong base. An equal number of moles of
each will create a neutral solution.
(.1500L)(.100M NaOH) = .01500 moles
NaOH, and therefore .015 moles OH-
.01500 moles H+ = .01500 moles HNO3 =
(.120M HNO3)(volume HNO3)
volume HNO3 = .125L HNO3, or 125mL
HNO3
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In a ________ change, such as a change of state, the substance as a whole changes, but its underlying structure remain the same.
pav-90 [236]

Answer:

Im pretty sure it is chemical change

4 0
2 years ago
Helen recorded the following data about the half-life of a radioisotope. Radioactive Decay of Radioisotope A Grams of Radioactiv
MissTica

Answer:

Line graph

Explanation:

I did it on Study Island

7 0
3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
2 years ago
Problem Page Question It takes to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbo
Marizza181 [45]

This is a incomplete question. The complete question is:

It takes 348 kJ/mol to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon. Round your answer to correct number of significant digits

Answer: 344 nm

Explanation:

E=\frac{Nhc}{\lambda}

E= energy  = 348kJ= 348000 J  (1kJ=1000J)

N = avogadro's number = 6.023\times 10^{23}

h = Planck's constant = 6.626\times 10^{-34}Js


c = speed of light = 3\times 10^8ms^{-1}

348000=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{348000}

\lambda=3.44\times 10^{-7}m=344nm    1nm=10^{-9}m

Thus the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon is 344 nm

5 0
2 years ago
A cork has a mass of 3 grams and a volume of 16 cms calculate the density
REY [17]

hey mate here is ur answer

solution

mass{m}=3 gram

             =3/1000

             

volume{v}=16cm

                =16/100

               

density=m/v

           =3/1000÷16/100

           =3/160

            =0.01875kg/m3

6 0
3 years ago
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