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3 years ago

Refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances. Assume all concentrations are 1 M.

Chemistry

2 answers:

LekaFEV [45]3 years ago

7 0

Answer:

c) HCl and NaCl

Explanation:

Since all the solutions are on a 1:1 mole ratio the comparison is straight forward.

The lowest pH will be for solution c) which has a strong acid, HCl, which ionizes 100 % and the neutral salt NaCl (which is neutral since it is derived from the reaction of the strong acid HCl and the strong base NaOH).

Solutions a) and b) are buffers of the weak base NH₃ and its conjugate acid NH₄⁺ and weak acid H₃PO₄ and its conjugate weak base NaH₂PO₄ respectively.

Solution c) is a basic solution being a mixture of the weak base NH₃ and the strong base NaOH

Solution e) is a mixture of a weak base NH₂ and weak acid HC₂H₃O₂

denis23 [38]3 years ago

7 0

Answer:

HCl and NaCl option c

Explanation:

According to the pH scale interpretation a lower pH means that its is more acidic, while a higher pH means that it is more basic.

Simply put a pH of 7 is neutral. A pH less than 7 is acidic. A pH greater than 7 is basic.

From the question above c fits in the option with the lower pH meaning its a strong acid and it tends more towards 0 than 7.

A solution, containing a buffer, is one that maintains a relatively constant pH.

NH3 and NH4Cl ;H3PO4 and NaH2PO4 are an acid/base conjugate pair each. They will make an excellent buffer.

NaOH and NH3 would rather make a slightly basic mixture

NH2 and HC2H3O2 (acetic acid) slightly basic mixture ..

Solution c has the lowest pH.

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To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o

Romashka-Z-Leto [24]

Answer:

a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

$\pi r^{2}h$ ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

$r=\frac{d}{2}$

$=\frac{3.36x10^{2}}{2}$

= 168 m

Volume =

$=\frac{22\times 168^{2}\times 21.80}{7}$

$=1.93\times 10^{6} m^{3}$

2.Convert ppm to g/m3 and Solve for mass of F-

$1ppm = 1g/m^{3}$

$0.9ppm = 0.9g/m^{3}$

Because both ppm and g/m3 are same quantity .

$g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}$

$0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}$

$mass\ of\ F- =1.740g$

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

$2Na+F_{2}\rightarrow 2NaF$

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

$\frac {84}{38}\times 1.74$

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

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3 years ago

A particular reactant decomposes with a half‑life of 109 s when its initial concentration is 0.280 M. The same reactant decompos

Sophie [7]

Answer:

The order of reaction is 2.

Rate constant is 0.0328 (M s)⁻¹

Explanation:

The rate of a reaction is inversely proportional to the time taken for the reaction.

As we are decreasing the concentration of the reactant the half life is increasing.

a) For zero order reaction: the half life is directly proportional to initial concentration of reactant

b) for first order reaction: the half life is independent of the initial concentration.

c) higher order reaction: The relation between half life and rate of reaction is:

Rate = $\frac{1}{k[A_{0}]^{(n-1)}}$

Half life = $K\frac{1}{[A_{0}]^{(n-1)} }$

$\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }$

where n = order of reaction

Putting values

$\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}$

$0.472=(0.472)^{(n-1)}$

Hence n = 2

$halflife=\frac{1}{k[A_{0}]}$

Putting values

$231=\frac{1}{K(0.132)}$

K = 0.0328

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Answer:

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