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makkiz [27]
3 years ago
11

People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of m

ilk of magnesia is 1 teaspoon, which contains 500 mg of Mg(OH)2. What volume of HCl solution with a pH of 1.25 can be neutralized by 1 dose of milk of magnesia
Chemistry
1 answer:
Llana [10]3 years ago
3 0

Answer:

V_{HCl}=0.208L=208mL

Explanation:

Hello,

In this case, since the chemical reaction is:

2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

n_{HCl}=2*n_{Mg(OH)_2}

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g}  =0.00858mol

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M

Then, since the concentration and the volume define the moles, we can write:

[HCl]*V_{HCl}=2*n_{Mg(OH)_2}

Therefore, the neutralized volume turns out:

V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL

Best regards.

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Tamoxifen is an irreversible, competitive inhibitor.

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
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Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

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Answer:

a) = 0.704%

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Explanation:

Given that:

K_a= 1.34*10^{-5

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Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

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