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3 years ago

An aqueous solution contains 0.23 M potassium hypochlorite.

Chemistry

1 answer:

Arturiano [62]3 years ago

5 0

Answer:

0.22 mol HClO, 0.11mol HBr.

0.25mol NH₄Cl, 0.12 mol HCl

Explanation:

A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.

Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of 0.22 mol HClO will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, 0.11mol HBr will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.

Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of 0.25mol NH₄Cl will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of 0.12 mol HCl will produce NH₄⁺. 0.25mol HCl consume all NH₃.

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The conversion of methyl isonitrile to acetonitrile in the gas phase at 250 °C CH3CN(g) is first order in CH3NC with a rate cons

nadya68 [22]

Answer: The concentration of $CH_3NC$ will be $1.56\times 10^{-2}M$ after 416 seconds have passed.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.00\times 10^{-3}s^{-1}$

t = age of sample = ?

a = let initial amount of the reactant = $5.45\times 10^{-2}M$

a - x = amount left after decay process = $1.56\times 10^{-2}M$

$t=\frac{2.303}{3.00\times 10^{-3}}\log\frac{5.45\times 10^{-2}}{1.56\times 10^{-2}}$

$t=416s$

The concentration of $CH_3NC$ will be $1.56\times 10^{-2}M$ after 416 seconds have passed.

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3 years ago

Which of the following are acids by the Bronsted-Lowry Definition:

olga55 [171]

ANSWER IS (A)

EXPLANATION:

Bronsted-Lowry concept states that a substance is an acid if it can act as a H+ donor.

HCl in aqueous solution means that HCl is present in water, HCl + H2O --> H3O+ + Cl-. This reaction will take place, the H+ from HCl will be donated to H2O. So, HCl is a bronsted-lowry acid by definition.

However, Methanol (CH3OH) its written that it is liquid, i.e. pure methanol, CH3OH(l). It is both acidic as well as basic. when it is mixed with water then it behaves as an acid.

The last one ammonia in gas phase is also neutral because its not in water. if mixed in water it behaves as a base.

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3 years ago

If a 200 g piece of aluminum has a density of 5.0 g/cm^3. what is its volume?

Strike441 [17]

Answer:

Volume=mass in g /density

4 0

2 years ago

Read 2 more answers

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Isotopes (such as hydrogen-1, hydrogen-2, and hydrogen-3) are atoms of the same element that differ in:

maria [59]

They have a different amount of neutrons.

8 0

3 years ago