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3 years ago

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What the difference of meaning between strong and weak Base. gave one example for each of them

1 answer:

kicyunya [14]3 years ago

8 0

1.Strong basses dissociates into ions almost 100%.

But weak basses partially dissociates into ions.

2.Strong basses has pH value smaller but closer to 14 and weak basses has pH value greater but cloaer to 7.

3.strong basses ate highly reactive whereas weak basses are less reactive.

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A student makes several measurements of the density of an unknown mineral sample. She then reports the average value of these me

Tomtit [17]

Answer:

B). Precision.

Explanation:

During the course of measurements, two important factors are been seen to play key part which are accuracy and also precision. Precision here can reflect to the closeness of an answer towards the main value even though accuracy can show that too. Also it can show series of values been reproduced in the measurement flow; though in some cases can be a bit far from the actual value. Also its values are swen to be different because of repeatition and alsio in some cases, observational errors too.

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3 years ago

30 mL of 0.25 M acetic acid are titrated with 0.05 M KOH. What is the pH after addition of 75 mL KOH? Group of answer choices

hjlf

Answer:

The PH of the mixture is 4.74

Explanation:

The number of millimoles of acetic acid is calculated using the formula:

No of millimoles= Molarity * Volume( in ml)

= 0.25M * 30ml = 7.5 moles

Number of millimoles of KOH is calculated using:

Number of millimoles = Molarity * Volume ( in ml)

=0.05M * 75ml

= 3.75 moles

The PH of the solution is derived using:

pH = pKa + log [salt] / acid

= $-log [ 1.8 * 10^5 ] + log [ 3.75 mmoles/ 3.75 mmoles]$

=4.74

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4 years ago

What is the mass, in grams, of a sample of 6.98 × 1024 atoms of magnesium (Mg)?

monitta

N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹

n(Mg)=N/Nₐ

m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ

m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g

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3 years ago

According to the wave model of light what makes blue light different from green light

34kurt

Answer: Green is 550 nm and blue is 470 nm

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3 years ago

Read 2 more answers

249 g of potassium iodide, KI, is mixed with 496.5 g of lead(II) nitrate, Pb(NO3)2.

sveticcg [70]

<u>Answer: </u>The number of moles of excess reagent left is 0.750 moles

<u>Explanation:</u>

The number of moles is calculated by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

<u>For KI:</u>

Given mass of $KI$ = 249 g

Molar mass of $KI$ = 166 g/mol

Putting values in equation 1, we get:

$\text{Moles of KI}=\frac{249g}{166g/mol}=1.5mol$

<u>For </u> $Pb(NO_3)_2$ <u>:</u>

Given mass of $Pb(NO_3)_2$ = 496.5 g

Molar mass of $Pb(NO_3)_2$ = 331 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Pb(NO_3)_2=\frac{496.5g}{331g/mol}=1.5mol$

The given chemical equation follows:

$2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2$

By stoichiometry of the reaction:

If 2 moles of KI reacts with 1 mole of lead(II) nitrate

So, 1.5 moles of KI will react with = $\frac{1}{2}\times 1.5=0.75mol$ of lead(II) nitrate

As the given amount of lead(II) nitrate is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, KI is considered a limiting reagent because it limits the formation of the product.

Moles of excess reactant ( $Pb(NO_3)_2$ ) left = [1.5 - 0.75] = 0.750 moles

Hence, the number of moles of excess reagent left is 0.750 moles

8 0

3 years ago