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1 year ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

Chemistry

1 answer:

kakasveta [241]1 year ago

7 0

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

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The concentration of C29H60 in summer rainwater is 34 ppb. Find the molarity of this compound in nanomoles per liter (nM).

Naya [18.7K]

Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter

Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.

The algebraic expression would be:

ppb [=] micrograms of compound/liter of solution

We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.

For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.

The respective procedure is in a attached file.

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1 year ago

A solution with a hydrogen ion concentration of 3.25 × 10-2 m is ________ and has a hydroxide concentration of _______

Romashka-Z-Leto [24]

To know the acidity of a solution, we calculate the pH value. The formula for pH is given as:

pH = - log [H+] where H+ must be in Molar

We are given that H+ = 3.25 × 10-2 M

Therefore the pH is:

pH = - log [3.25 × 10-2]

pH = 1.488

Since pH is way below 7, therefore the solution is acidic.

To find for the OH- concentration, we must remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,

[H+]*[OH-] = 10^-14 
[OH-] = 10^-14 / [H+]

[OH-] = 10^-14 / 3.25 × 10-2

[OH-] = 3.08 × 10-13 M

Answers:

Acidic

[OH-] = 3.08 × 10-13 M

6 0

1 year ago

PLEASE HELP!!!!!!!!!!!!!

lions [1.4K]

Answer:

Pd(O2C2H3)2

Explanation:

Percentage composition of elements in the compound:

Pd= 47.40%

O=28.50%

C=21.40%

H= 2.69%

Molar mass of each element

Pd= 106g/mol

O= 16g/mol

C= 12g/mol

H= 1g/mol

Step 1

Find the moles of each element in 100g sample using the formula

no. of moles(n)=mass/molar mass

For Pd: 47.40g/106g/mol

n= 0.447 moles

For O: 28.50/16

n= 1.78 moles

For C: 21.40/12

n= 1.783 moles

For H: 2.69/1

n= 2.69moles

Step 2

Now divide the no. of moles separately by the smallest number of moles. Smallest number of moles is 0.447

Pd= 0.447/0.447

= 1

O= 1.78/0.447

= 3.9=4

C= 1.78/0.447

= 3.9=4

H= 2.69/0.447

= 6

Hence empirical formula is PdO4C4H6

The probable molecular formula is $Pd(O_{2}C_{2}H_{3})2$

8 0

1 year ago

Read 2 more answers

A 1000.0 ml sample of lake water in titrated using 0.100 ml of a 0.100 M base solution. What is the molarity (M) of the acid in

Fittoniya [83]

The molarity (M) of the acid in the lake water is $0.00001M$ .

Explanation:

In order to estimate the concentration of a solution in molarity, then the total number of moles of the solute is divided by the total volume of the solution.

According to the given information, the formula will be applied for calculating molarity (M) of the acid in the lake water is :

$M_1V_1=M_2V_2$

Here;

$M_1,M_2$ are molarity of acid in the lake water and base solution respectively. $V_1,V_2$ are volume of sample in the lake water and base solution respectively.

Given values are as follows:

$M_1=?\\M_2=0.100M\\V_1=1000ml\\V_2=0.100ml$

Putting these values in above equation :

$M_1V_1=M_2V_2$

$M_1(1000)=(0.100)(0.100)$

$M_1=\frac{(0.100)(0.100)}{1000}$

$M_1=0.00001M$

Therefore, the molarity (M) of the acid in the lake water is $0.00001M$ .

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1 year ago

A readily corroded silvery solid that fizzes with water

fiasKO [112]

Answer:

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Explanation:

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