The number of _______ in an atom of an element is the element's atomic number.

Calcule el poh de una disolucion de acido bromhidrico hbr 0.176m

Lerok [7]

Answer:

pOH = 13.246

Explanation:

pH = -log[H+]

[H+] = [HBr] porque HBr es un acido fuerte.

pH = -log(0.176) = 0.754

pH + pOH = 14

pOH = 14 - 0.754 = 13.246

4 0

2 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can calculate the pOH (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can calculate the pH value:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

8 0

3 years ago

Carry out the following calculation, paying special attention to the significant figures (where 4/3 is exact), rounding, and uni

babunello [35]

Answer:

Value = 1.80 g/cm³ (Approx)

Explanation:

Given:

$\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3}$

Computation:

$\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3} \\\\\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)} \\\\ \frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)}\\\\ \frac{3.39 \times 10^7g}{18.8166132\times 10^6 cm^3} \\\\ 1.80159945g/cm^3$

Value = 1.80 g/cm³ (Approx)

7 0

3 years ago

Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer

bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0

2 years ago

Click oil basics to explore the natural resources needed for the production of plastics. use the information to answer the quest

GaryK [48]

The ingredients that is important in making plastics is natural gas plant liquids.

Natural gas liquids (NGLs) are components of natural gas which are separated from the gas state in the form of liquids. They many applications including; plastic production, inputs for petrochemical plants, they are also burned for space heat and cooking, and can also be blended into vehicle fuel.

In the manufacturing of plastics, come components of NGLs are used. And for this case, ethane. Ethane is used in the production of ethylene which are passed through pressure and catalyst to be turned to plastics like polythene.

The reason why they are mostly used in the manufacturing plastics than other resources like crude oil is that it is much cleaner.

Learn more about natural gas liquids here:

brainly.com/question/20415322

#SPJ4

8 0

1 year ago