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alexira [117]
4 years ago
10

What is the ph of a soft drink in which the major buffer ingredients are 6.6 g of nah2po4 and 8.0 g of na2hpo4 per 355 ml of sol

ution?
Chemistry
1 answer:
Alex777 [14]4 years ago
5 0
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
                                                   = 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
  = 0.055/ ( 355 ×10^-3)
  = 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)

1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
                                             = 0.0563 moles
 [HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
     = 0.0563/(355×10^-3)
     =  0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation 
Ka of H2PO4- = 6.20 × 10^-8
 [H+] =Ka*([H2PO4-]/[HPO4(2-)]
        = (6.20 ×10^-8)×(0.155/0.1586)
        = 6.059 ×10^-8 M
pH = - log[H+]
     = - log (6.059×10^-8)
     = 7.218

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Which of the possible compounds has a mass of 163 grams when
Simora [160]

Answer:

CH4

Explanation:

In solving this problem, we must remember that one mole of a compound contains Avogadro's number of elementary entities. These elementary entities include atoms, molecules, ions etc. Recall that one mole of a substance is the amount of substance that contains the same number of elementary entities as 12g of carbon-12. The Avogadro's number is 6.02 × 10^23.

Hence we can now say;

If 163 g of the compound contains 6.13 ×10^24 molecules

x g will contain 6.02 × 10^23 molecules

x= 163 × 6.02 × 10^23 / 6.13 × 10^24

x= 981.26 × 10^23/ 6.13 ×10^24

x= 160.1 × 10^-1 g

x= 16.01 g

x= 16 g(approximately)

16 g is the molecular mass of methane hence x must be methane (CH4)

6 0
3 years ago
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described b
Katyanochek1 [597]

Answer:

m_{MnO_2}=1.378gMnO_2

Explanation:

Hello,

In this case, we first apply the ideal gas equation to compute the moles of produced chlorine:

n_{Cl_2}=\frac{PV}{RT}=\frac{765 Torr*\frac{1atm}{760Torr}*0.385L}{0.082\frac{atm*L}{mol*K}*298.15K} =0.01585molCl_2

Then, by considering the given reaction, applying the stoichiometry, that shows a 1 to 1 relationship between chlorine and manganese dioxide, we find:

m_{MnO_2}=0.01585molCl_2*\frac{1molMnO_2}{1molCl_2} *\frac{86.937gMnO_2}{1molMnO_2} \\m_{MnO_2}=1.378gMnO_2

Best regards.

3 0
3 years ago
A sample of gas has a mass of 38.8 mg. Its volume is 224 mL at a temperature of 55 °C and a pressure of 886 torr. Find the molar
Pie

Answer:

4g/mol

Explanation:

Firstly, we can get the number of moles of the gas present using the ideal gas equation.

PV = nRT

Here:

P = 886 torr

V = 224ml = 224/1000 = 0.224L

T = 55 degrees celcius= 55+ 273.15 = 328.15K

R = molar gas constant = 62.36 L⋅Torr⋅K−1⋅mol−1

n = PV/RT

n = (886 * 0.224)/(62.36 * 328.15)

n = 0.009698469964 mole

Now to get the molar mass, this is mathematically equal to the mass divided by the number of moles. We have the mass and the number of moles, remaining only the molar mass.

First, we convert the mass to g and that is 38.8/1000 = 0.0388

The molar mass is thus 0.0388/0.009698469964 = 4g/mol

3 0
4 years ago
Pls help 2-3 tyvm ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎
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Answer:

I could I guess?

But like only 1 lol

Explanation:

6 0
3 years ago
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The daughter nucleus that is produced is ₂₉⁶³cu.

What is Electron capture?

The process of drawing an electron to the nucleus, where it combines with a proton to create a neutron and a neutrino particle, is known as electron capture.

The daughter nucleus is the nucleus that is made by the parent nucleus. The nucleus that remains after the decay is called the daughter nucleus.

Here is the chemical formula for the reaction that results in the electron capture of the zinc-63 nucleus:

_A^ZX + e^- =  _A^Z_-_1 Y + ye

_3_0^6^3Zn\; + e^- = _2_9^6^3Cu + ye

Thus, the daughter nucleus produced when zn63 undergoes electron capture is _2_9^6^3Cu.

To learn more about Electron capture, refer to the below link:

brainly.com/question/10964824

#SPJ4

4 0
2 years ago
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