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3 years ago

Determine the density of nh3 gas at 435k and 1.00atm

1 answer:

5 0

We assume that this gas is ideal. Therefore, we can use the ideal gas equation which is expressed as:

PV=nRT

We manipulate this equation to give us an expression which will correspond to density. We do as follows:

PV= nRT
P/RT = n/V where n = m/MM
P(MM) /RT = m/V = density
Density = 1.00 (17.03) / 0.08206 (435)
Density = 0.48 g / L

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Question 2 Calculate the pH of a solution that has an acetic acid concentration of 0.05 M and a sodium acetate concentration of

Eva8 [605]

Answer : The pH of the solution is, 4.9

Explanation : Given,

Dissociation constant for acetic acid = $K_a=1.8\times 10^{-5}$

Concentration of acetic acid = 0.05 M

Concentration of sodium acetate = 0.075 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.8\times 10^{-5})$

$pK_a=5-\log (1.8)$

$pK_a=4.7$

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}$

Now put all the given values in this expression, we get:

$pH=4.7+\log (\frac{0.075}{0.05})$

$pH=4.9$

Therefore, the pH of the solution is 4.9.

8 0

3 years ago

A 25.0 ml sample of 0.723 m hclo4 is titrated with a 0.273 m koh solution. the h3o concentration after the addition of 50.0 ml o

3241004551 [841]

This is an acid base reaction and the chemical equation for the above reaction is as follows;
KOH + HClO₄ ---> KClO₄ + H₂O
the stoichiometry of acid to base is 1:1
KOH is a strong base and HClO₄ is a strong acid therefore they both ionize completely into their respective ions
Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol
Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol
since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base.
The excess base remaining is - 0.018 - 0.013 = 0.005 mol
total volume of solution = 25.0 mL + 50.0 mL = 75.0 mL
[OH⁻] = 0.005 mol/0.075 L = 0.067 M
pOH = -log[OH⁻]
pOH = -log(0.067 M)
pOH = 1.17
pOH + pH = 14
Therefore pH = 14 - 1.17 = 12.83
by knowing pH we can calculate the [H₃O⁺]
pH = -log [H₃O⁺]
[H₃O⁺] = antilog[-12.83]
[H₃O⁺]= 1.47 x 10⁻¹³ M

5 0

3 years ago

What happens when sugar is heated? Give a balanced equation

mario62 [17]

Answer:

when the sugar is heated it turn out in caramelize

8 0

3 years ago

Our blood which is made up of different types of cells is classified as what level of organization? *

beks73 [17]

Answer:

c.tissue