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3 years ago

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Chemistry

1 answer:

lapo4ka [179]3 years ago

5 0

1. Liquid
2. Gas
3. Solid

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g Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution.

jarptica [38.1K]

Answer:

Following are the responses to the given choices:

Explanation:

For point a:

Using the acid and base which are strong so,

moles of $H^+$ (from $HNO_3$ )

$= 24.9\ mL \times 0.100\ M \\\\= \frac{24.9}{1000\ L} \times 0.100\ M \\\\= 2.49 \times 10^{-3} \ mol$

moles of $OH^{-}$ (from $KOH$ )

$= 25.0\ mL \times 0.100\ M \\\\= \frac{25.0}{1000 \ L} \times 0.100 \ M \\\\\= 2.50 \times 10^{-3}\ mol$

$1\ mol H^{+} \ neutralizes\ 1\ mol\ of\ OH^{-}$

So, $(2.50 \times 10^{-3} mol - 2.49 \times 10^{-3} mol)$ i.e. $1 \times 10^{-5}$ mol of $OH^-$ in excess in total volume $(24.9+25.0) \ mL = 49.9 \ mL$ i.e. concentration of $OH^- = 2 \times 10^{-4}\ M$

$p[OH^{-}] = -\log [OH^{-}] = -\log [2 \times 10^{-4}\ mol] = 3.70$

Since, $pH + pOH = 14,$

so,

$\to pH = 14- pOH = 14- 3.70 = 10.30$

For point b:

moles of $OH^-$ = from point a $= 2.50 \times 10^{-3} \ mol$

moles of $H^+$ (from $HNO_3$ ):

$= 25.1 mL \times 0.100 M\\\\ = \frac{25.1}{1000}\ L \times 0.100 \ M\\\\ = 2.51\times 10^{-3} \ mol$

1 mol $H^+$ neutralizes 1 mol of $OH^-$

So, $(2.51 \times 10^{-3}\ mol - 2.50 \times 10^{-3}\ mol)$ i.e. $1 \times 10^{-5} \ mol \ of\ H^+$ in excess in the total volume of $(25.1+25.0) \ mL = 50.1\ mL$ i.e. concentration of $H^+ = 2 \times 10^{-4}\ M$

Hence, $pH = -\log [H^+] = -\log[2 \times 10^{-4}] = 3.70$

6 0

3 years ago

What must the value of a conversion factor always be?

storchak [24]

Answer:

One or unity or 1

Explanation:

For example,

1 min =60 s

If we divide each side of the equation by 1 min, we get

$1 = \dfrac{\text{60 s}}{\text{1 min}}$

If we divide each side of the equation by 60 s, we get

$\dfrac{\text{1 min}}{\text{60 s}} = 1\\\\\text{Thus, both conversion factors, $\dfrac{\text{60 s}}{\text{1 min}}$ and $\dfrac{\text{1 min}}{\text{60 s}}$, equal 1.}$

8 0

3 years ago

over a 12.3 minuete period 5.13 E-3 moles of F2 gas effuses from a contaier. How many moles of CH4 gas could effuse from from th

Aleonysh [2.5K]

Answer : The moles of methane gas could be, $7.90\times 10^{-3}mol$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$

$[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}$

where,

$R_1$ = rate of effusion of fluorine gas

$R_2$ = rate of effusion of methane gas

$n_1$ = moles of fluorine gas = $5.13\times 10^{-3}mol$

$n_2$ = moles of methane gas = ?

$t_1=t_2$ = time = 12.3 min (as per question)

$M_1$ = molar mass of fluorine gas = 38 g/mole

$M_2$ = molar mass of methane gas = 16 g/mole

Now put all the given values in the above formula 1, we get:

$[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}$

$n_2=7.90\times 10^{-3}mol$

Therefore, the moles of methane gas could be, $7.90\times 10^{-3}mol$

8 0

3 years ago

Drinking water should have sufficiently low levels of __________ to prevent disease. What is the missing word in this sentence?

ELEN [110]

For water to be drinkable, it must have sufficiently low levels of dissolved salts and microbes.

3 0

3 years ago

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

3 years ago