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Mekhanik [1.2K]
3 years ago
7

a chemist wishes to mix some pure acid with some water to produce 16L of a solution that is 30% acid how much pure acid and how

much water should be mixed?
Chemistry
1 answer:
Aloiza [94]3 years ago
4 0

<u>Answer:</u> The volume of acid and water that must be mixed will be 4.8 L and 11.2 L

<u>Explanation:</u>

We are given:

Volume of mixture = 16 L

Percent of acid present = 30 %

Calculating the percentage of acid present in the mixture:

\Rightarrow 16\times \frac{30}{100}=4.8L

The mixture is made entirely of acid and water.

Volume of acid in the mixture = 4.8 L

Volume of water in the mixture = 16 - 4.8 = 11.2 L

Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L

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How is an oxidation half-reaction written using the reduction potential chart? How is the oxidation potential voltage determined
insens350 [35]
Oxidation  half  reaction  is  written   as    follows when  using  using  reduction potential  chart
example  when  using  copper  it is  written  as  follows
CU2+   +2e-  --> c(s)  +0.34v
 oxidasation  is the  loos  of  electron  hence copper oxidation  potential  is  as  follows
cu (s)  -->  CU2+   +2e    -0.34v

7 0
3 years ago
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pishuonlain [190]
Some of the NH₄+ will combine with the OH- and shift the equilibrium backwards and from NH₄OH to balance the change produced by addition of NH₄+ ions.
5 0
3 years ago
How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit
NARA [144]
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
5 0
3 years ago
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A sample of nitrogen gas is produced in a reaction and collected under water in a graduated cyliner. The temperature is 26.3 oC
Rudiy27

The mass of nitrogen collected is mathematically given as

M-N2=0.025gram

<h3>What is the mass of nitrogen collected?</h3>

Question Parameters:

A sample weighing 2.000g

the liberated NH3 is caught in  50ml pipeful  of H2SO4 (1.000ml   =  0.01860g Na2O).

T=26.3c=299.3K

Pressure=745mmHg=745torr

Pressure of N2=745-25.2=719.8torr

Generally, the equation for the ideal gas   is mathematically given as

PV=nRT

Therefore

719.8/760=45.6/1000=n*0.0821*299.3

n=0.00176*14

In conclusion, the Mass of N2

M-N2=0.00176*14

M-N2=0.025gram

Read more about Mass

brainly.com/question/4931057

8 0
2 years ago
What is the poH of a<br> 2.6 x 10-6 M H+ solution?
andriy [413]

Answer:

Explanation:

pH and pOH.....

The pH is a way of expressing the hydrogen ion concentration.

pH = -log[H+] ............. where [x] means "the concentration of x in moles per liter."

From pH you can compute pOH since at 25C pH + pOH = 14.00 .......... (but only at 25C)

pH = -log(2.6x10^-6) = 5.585 ..... which should be rounded to two significant digits: pH = 5.59

When taking the log of a number, only the digits to the right of the decimal reflect the precision in the original number. Since 2.6x10^-6 has two significant digits, a pH of 5.59 has two significant digits.

pOH + pH = 14.00

pOH = 14.00 - pH = 14.00 - 5.59 = 8.41 ......... at 25C

We can also use the H+ ion concentration to get the hydroxide ion concentration and from that the pOH.

Kw = [H+][OH-] = 1.00x10^-14 .......... at 25C .... like any Kc, the value changes with temperature

[OH-] = Kw / [H+] = 1.00x10^-14 / 2.6x10^-6 = 3.846x10^-9 .... to a couple of guard digits

pOH = -log[OH-] = -log(3.846x10^-9) = 8.415 ...... round to two significant digits: pOH = 8.42 ..... at 25C

=========

Just for grins, you might want to know how Kw changes with temperature, and how [H+] and [OH-] are related at some other temperatures. The pH is the pH of a neutral solution at various temperatures. For instance at 10C a neutral solution has a pH of 7.27. That's not a basic pH. 7.27 is the pH of a neutral solution, but at a different temperature. In a neutral solution at 10C [H+] = [OH-] = 5.41x10^-8M.

pH and Kw for a neutral solution at different temperatures

T .........pH ......... Kw

0......... 7.47....... 0.114 x 10-14

10....... 7.27....... 0.293 x 10-14

20....... 7.08....... 0.681 x 10-14

25....... 7.00....... 1.008 x 10-14

30....... 6.92....... 1.471 x 10-14

40....... 6.77....... 2.916 x 10-14

50....... 6.63....... 5.476 x 10-14

100..... 6.14....... 51.3 x 10-14

4 0
3 years ago
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