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2 years ago

What is the best definition of weight

Chemistry

2 answers:

LenKa [72]2 years ago

7 0

Answer:

a body's relative mass or the quantity of matter contained by it, giving rise to a downward force; the heaviness of a person or thing.

really depending on what weight you are referring to but thats the number 1 on dictonary

enyata [817]2 years ago

5 0

Somethings mass or the matter that is held inside of an object

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Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$

From the reaction, we conclude that

As, 1 mole of $KI$ react to give 1 mole of $AgI$

So, 0.0178 moles of $KI$ react to give 0.0178 moles of $AgI$

Thus,

Moles of AgI = Moles of $I^-$ anion = Moles of $Ag^+$ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

$\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}$

$\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M$

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0

3 years ago

Using a ph meter, you find the ph of an unknown solution to be 8. 0. How would you describe this solution?.

insens350 [35]

Answer:

Just Barely Base/Neutral

Explanation:

a pH of 8.0 is greater than Neutral (7.0) but is still neutral due to it being more neutral than a base

4 0

2 years ago

What does AuCl, Gold (I) Chloride, decompose into?

Alexxx [7]

Answer:

Gold

Explanation:

Gold(I) chloride is a compound of gold and chlorine with the chemical formula AuCl

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3 years ago

6 CO 2 + 12 H 2 O C 6 H 12 O 6 + 6 H 2 O + 6 O 2

adelina 88 [10]

During photosynthesis plant release
<span>C 6 H 12 O 6 + 6 H 2 O + 6 O 2
the gas is 6O2
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3 years ago

A chemist dissolves 248mg of pure hydrochloric acid in enough water to make up 140mL of solution.Calculate the pH of the solutio

evablogger [386]

Answer : The pH of the solution is, 1.31

Explanation : Given,

Mass of $HCl$ = 248 mg = 0.248 g (1 mg = 0.001 g)

Volume of solution = 140 mL

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the concentration of hydrochloric acid.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }HCl\times 1000}{\text{Molar mass of }HCl\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{0.248g\times 1000}{36.5g/mole\times 140mL}=0.0485mole/L=0.0485M$

The concentration of HCl is, 0.0485 M

As we know that, HCl dissociates to give hydrogen ion and chloride ion.

The dissociation reaction is:

$HCl\rightarrow H^++Cl^-$

Concentration of HCl = Concentration of H⁺ = Concentration of Cl⁻ = 0.0485 M

Now we have to calculate the pH of the solution.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

Mathematically,

$pH=-\log [H^+]$

$pH=-\log (0.0485)$

$pH=1.31$

Therefore, the pH of the solution is, 1.31

6 0

3 years ago

What is the best definition of weight​

What is the best definition of weight