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3 years ago

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How many molecules of ammonia are contained in 10.4 moles of ammonia, NH3?

1 answer:

seraphim [82]3 years ago

4 0

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We do as follows:

10 mol NH3 ( 6.022 x 10^23 molecules / 1 mol ) = 6.022x10^24 molecules NH3

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A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

number of moles of $NH_4Cl$ = 0.40 mol/L × 200 × 10⁻³L

= 0.08 mole

number of moles of $NaOH$ = 0.80 mol/L × 50 × 10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

Initial (M) 0.08 0.04 0

Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

$K_a*K_b = 10^{-14} \ at \ 25^0C$

$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

$K_a= 5.68*10^{10}$

$pK_a = - log \ (K_a)$

$pK_a = - log \ (5.68*10^{-10})$

$pK_a = 9.25$

$pH = pKa + \ log (\frac{HB}{HA} )$ for buffer solutions

$pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} )$ since they are in the same solution

$pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )$

$pH = 9.25$

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weqwewe [10]

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valina [46]

2)c

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Convert 0.00200 g into Mg and answer in scientific notation:

Masja [62]

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$0.00200g=\frac{1000mg}{1g}$

Since you go straight across its just multiplication so 0.00200x1000=200mg

To answer it in scientific notation it should just be $2.00x10^{2}$ mg

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TEA [102]

Answer:

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