Which is a molecule?<br> A. Ne<br> B. NaCL<br> C. Trail Mix<br> D. graphene

What is the oxidation state of phosphorus in the following compounds?

Korolek [52]

A) H3PO4
H(+1),
P(X)
O(-2)
3*(+1)+x+4*(-2)=0, 3+x-8=0, x=+5
b)P2O5
P(x)
O(-2)
2x-2*5=0, 2x=10, x=+5
c) PH3
P(x)
H(+1)
x+3*(+1)=0, x+3=0, x=-3

5 0

4 years ago

A 0.2 M carboxylic acid (RCOOH) has a Ka = 1.66x10-6. What is the pH of this solution? Enter to 2 decimal places.

maria [59]

Answer:

3.24

Explanation:

The dissociation equation for the carboxylic acid can be represented as follows:

RCOOH —-> RCOO- + H+

We can use an ICE table to get the value of the concentration of the hydrogen ion. ICE stands for initial, change and equilibrium.

RCOOH RCOO- H+

Initial 0.2 0.0. 0.0

Change -x +x. +x

Equilibrium 0.2-x. x. x

We can now find the value of x as follows:

Ka = [RCOO-][H+]/[RCOOH]

(1.66* 10^-6) = (x * x)/(0.2-x)

(1.66 * 10^-6) (0.2-x) = x^2

x^2 = (3.32* 10^-7) - (1.66*10^-6)x

x^2 + (1.66 * 10^-6)x - (3.32* 10^-7) = 0

Solving the quadratic equation to get x:

x = 0.0005753650094369094 or - 0.0005753650094369094

As concentration cannot be negative, we discard the negative answer

Hence [H+] = 0.0005753650094369094

By definition, pH = -log[H+]

pH = -log(0.0005753650094369094)

pH = 3.24

8 0

3 years ago

Determine the percent ionization of a 0.230 m solution of benzoic acid.

Fudgin [204]

We need the dissociation constant of benzoic acid which is 6.3x10^'5. Then using the dissociation formula, ka = x2 / (Mo - x) where Mo is the initial concentration. x is determined then. percent ionization is computed as (x/Mo)* 100%. This is then the final answer.

5 0

4 years ago

Where else have you heard the term conversation? How does that term’s meaning in that other context relate to its meaning in che

Mrac [35]

I have no skins for you my brother and my girlfriend and my uncle are going out and have you come on the house for me to do you have to get on your phone

7 0

4 years ago

A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction go

4vir4ik [10]

Answer : The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of $NH_4Cl$ = 26.5 g

Mass of $NaOH$ = 10 g

Molar mass of $NH_4Cl$ = 53.5 g/mole

Molar mass of $NaOH$ = 40 g/mole

First we have to calculate the moles of $NH_4Cl$ and $NaOH$ .

$\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles$

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NH_4Cl+NaOH\rightarrow NH_4OH+NaCl$

From the balanced reaction we conclude that

As, 1 mole of $NaOH$ react with 1 mole of $NH_4Cl$

So, 0.25 moles of $NaOH$ react with 0.25 moles of $NH_4Cl$

From this we conclude that, $NH_4Cl$ is an excess reagent because the given moles are greater than the required moles and $NaOH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of $NH_4Cl$ .

$\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl$

$\text{Mass of }NH_4Cl=(0.245mole)\times (53.5g/mole)=13.1075g$

Therefore, the amount of reactant left in excess is 13.1075 grams.

5 0

3 years ago

Which is a molecule? A. Ne B. NaCL C. Trail Mix D. graphene