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ehidna [41]
3 years ago
6

How many Milligrams of potassium chloride KCl are required to make a 15 mL of a 1.20 M solution?​

Chemistry
1 answer:
Novay_Z [31]3 years ago
5 0

Answer:

1340 mg

Explanation:

concentration is the number of moles of solute in a 1000 mL of solution

the concentration of KCl solution is 1.20 M

There are 1.20 mol in 1000 mL of solution

a volume of 1000 mL contains - 1.20 mol

therefore 15 mL solution contains - 1.20 mol / 1000 mL x 15 mL = 0.018 mol

the molar mass of KCl is - 74.5 g/mol

mass of 1 mol is - 74.5 g

therefore mass of 0.018 mol is - 74.5 g/mol x 0.018 mol = 1.34 g

1000 mg - 1 g

therefore 1.34 g in milligrams is - 1.34 g x 1000 mg/g = 1340 mg

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Answer:

Ok:

Explanation:

So, you can use the Henderson-Hasselbalch equation for this:

pH = pKa + log(A^-/HA) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.

We can solve that

1 = log(A^-/HA\\) and so 10 = A^-/HA or 10HA = A-.  For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

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3 years ago
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3 years ago
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This is called as hydrochloric acid or hydrogen chloride.

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The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,
Masja [62]

<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.50-0=1.50V

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

where,

E_{cell} = electrode potential of the cell = 1.23 V

E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

Hence, the concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

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Korvikt [17]
Frequency = velocity (aka speed) / wave length
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