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2 years ago

1.

Chemistry

1 answer:

german2 years ago

3 0

Order of metals from least reactive to most reactive: B <C <A <D

<h3>Further explanation</h3>

Reducing agents are substances that experience oxidation

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction

Let's analyze the statement in the problem

I. Only A, C and D react with 1 mol/L HCl to give H₂(e)

M + HCl ⇒ MCl + H₂(MCl : alkali, MCl₂ : alkaline earth)

A, C and D can react with 1 mol / L HCl, meaning metals A, C and D are located to the left of element H (more reactive), and B in the right of element H

II. When A is added to solutions of the other metal ions, metallic B and C are formed but not D.

This means that metal A is more reactive than metals B and C, while D is more reactive than A, so metal D is the most reactive

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What volume of 1.10 M SrCl2 is needed to prepare 525 mL of 5.00 mM SrCl2?

grigory [225]

Answer:- 2.39 mL are required.

Solution:- It's a dilution problem and to solve this type of problems we use the dilution equation:

$M_1V_1=M_2V_2$

Where, $M_1$ and $M_2$ are molarities of concentrated and diluted solutions and $V_1$ and $V_2$ are their respective volumes.

$M_1$ = 1.10M

$M_2$ = 5.00mM = 0.005M (since, mM stands for milli molar and M stands for molar. 1M = 1000mM)

$V_1$ = ?

$V_2$ = 525 mL

Let's plug in the given values in the formula:

$1.10M(V_1)=0.005M(525mL)$

$V_1=(\frac{0.005M*525mL}{1.10M})$

$V_1=2.39mL$

So, 2.39 mL of 1.10M are needed to make 525 mL of 5.00mM solution.

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3 years ago

The atom in a sample of an element must contain nuclei with the same number of?

slava [35]

Hey there!

Answer: Protons

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3 years ago

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2 years ago

Read 2 more answers

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 $AgI$