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3 years ago

6

WILL MARK AS BRAINLIEST! PLEASE HELP

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RUDIKE [14]3 years ago

3 0

Answer:

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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 <u>reaction</u> between HCl and KOH:

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

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

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

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 <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> 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 <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

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

$60mL=0.06L$

Now we can calculate the concentration:

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

$M=0.0705$

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can <u>calculate the pH value</u>:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

8 0

3 years ago

Nataliya [291]

The daughter isotope : Radon-222 (Rn-222).

<h3>Further explanation</h3>

Given

Radium (Ra-226) undergoes an alpha decay

Required

The daughter nuclide

Solution

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,

alpha α particles ₂He⁴
beta β ₋₁e⁰ particles
gamma particles ₀γ⁰
positron particles ₁e⁰
neutron ₀n¹

The decay reaction uses the principle: the sum of the atomic number and mass number before and after decay are the same

Radium (Ra-226) : ₈₈²²⁶Ra

Alpha particles : ₂⁴He

So Radon-226 emits alpha α particles ₂He⁴ , so the atomic number decreases by 2, mass number decreases by 4

The reaction :

₈₈²²⁶Ra ⇒ ₂⁴He + ₈₆²²²Rn

4 0

3 years ago

Which of the following is true about most predators? (Select all that apply.)

-Dominant- [34]

Answer:

select all of them except they are the biggest

3 0

3 years ago

The electronic configuration of an element is given below.

drek231 [11]

Answer:

C. It is reactive because it must gain two electrons to have a full outermost energy level.

Step-by-step explanation:

The electron configuration of this element ends in 2s²2p⁴.

A filled energy level would end in 2s²2p⁶.

The element will be reactive, because it must gain electrons to have a full energy level, and it needs two more electrons to do this.

4 0

3 years ago

How many grams of carbonic acid were produced by the 3.00g sample of NaHCO3

Alexxx [7]

I dont know what subject is this

8 0

4 years ago