Which type of radiation is blocked by the skin but is the most dangerous once it is inside the body?

Find the pH. What are the pH values for the following solutions? (a) 0.1 M HCl (b) 0.1 M NaOH (c) 0.05 M HCl (d) 0.05 M NaOH

slega [8]

Answer:

(a) $pH=1$

(b) $pH=1.3$

(c) $pH=13$

(d) $pH=12.7$

Explanation:

Hello,

In this case, we define the pH in terms of the concentration of hydronium ions as:

$pH=-log([H^+])$

Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:

(a)

$[H^+]=[HCl]=0.1M$

$pH=-log(0.1)=1$

(b)

$[H^+]=[HCl]=0.05M$

$pH=-log(0.05)=1.3$

Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:

$[OH^-]=[NaOH]$

$pOH=-log([OH^-])$

$pH=14-pOH$

Thus, we have:

(b)

$pOH=-log(0.1)=1\\pH=14-1=13$

(d)

$pOH=-log(0.05)=1.3\\pH=14-1.3=12.7$

Best regards.

5 0

4 years ago

An atom has a diameter of approximately 0.0010 um. How many meters is this?

Likurg_2 [28]

0.0010 micrometer = 0.001 x 10^-6 meter = 10^-9 meter = 0.00000001 meter
.
Confusions ?

7 0

3 years ago

Hey I'm Chloe can you Help me, Thank you :)<br> How does photosynthesis use solar energy?

harina [27]

When a plant goes through photosynthesis, it uses light energy and converts it into chemical energy

4 0

3 years ago

Read 2 more answers

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of nh3 to no. 4 nh3(g) + 5 o2

ad-work [718]

Answer:- (a) $O_2$ , (b) 1.73 g NO and 1.55 g $H_2O$ , (c) 0.932 g of $NH_3$ , (d) yes, the results are consistent with law of conservation of mass.

Solution:- The given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

From balanced equation, ammonia and oxygen react in 4:5 mole ratio means 4 moles of ammonia reacts with 5 moles of oxygen to give the products.

(a) Let's calculate the moles of the reactants on dividing the grams by molar masses as:

For ammonia, $1.90gNH_3(\frac{1mole}{17g})$ = 0.112 mole

For oxygen, $2.30gO_2(\frac{1mole}{32g})$ = 0.0719 mole

From mole ratio, we need more moles of oxygen as compared to oxygen but from calculations the moles of oxygen are less than ammonia. So, oxygen is limiting reactant.

(b) To calculate the grams of each product formed we would calculate their moles first from limiting reactant that is oxygen and then multiply by their respective molar masses to convert the moles to grams as:

For grams of NO:

$0.0719moleO_2(\frac{4moleNO}{5moleO_2})(\frac{30gNO}{1moleNO})$ = 1.73 g of NO

For grams of $H_2O$ :

$0.0719moleO_2(\frac{6moleH_2O}{5moleO_2})(\frac{18gH_2O}{1moleH_2O})$ = 1.55 g of $H_2O$

So, 1.73 grams of NO and 1.55 grams of $H_2O$ would formed.

(c) Let's calculate the moles of excess reactant that is ammonia used to react with limiting reactant, oxygen as:

$0.0719moleO_2(\frac{4moleNH_3}{5moleO_2})$

= 0.0572 mole $NH_3$

Excess moles of ammonia = 0.112 - 0.0572 = 0.0548 mole

Now let's convert these excess moles of ammonia to grams on multiplying by molar mass:

$0.0548moleNH_3(\frac{17g}{1mole})$

= 0.932 g $NH_3$

So, after reaction when all the oxygen is used then 0.932 grams of ammonia would still be remaining.

(d) mass of reactants = mass of products + remaining mass of excess reactant

Let's plug in the values in it:

mass of reactants = 1.90 g + 2.30 g = 4.2 g

mass of products + remaining mass of excess reactant = 1.73 g + 1.55 g + 0.932 g

= 4.2 g

So, the results are consistent with law of conservation of mass.

3 0

4 years ago

Predict the neutral organic product of the following reaction. Include hydrogen atoms in your structure. When drawing hydrogen a

lukranit [14]

Answer:

The neutral organic product of the reaction is the 2-methoxy-2-methylpropane, shown in the attached figure.

Explanation:

By reacting an alkene (2-methylprop-1-ene) with an alcohol (methanol) in the presence of an acid it forms an ester (2-methoxy-2-methylpropane) through Markovnikov’s rule. First, a protonation of the alkene occurs where a tertiary carbocation is formed, then the nucleophile (methanol) attacks the previously formed carbocation, and finally ocurr the deprotonation of the hydrogen bound to the oxygen, thus forming the ester, in our case the neutral organic product 2-methoxy-2-methylpropane.

5 0

4 years ago