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3 years ago

Where do expect to find the most reactive metals on the periodic table;?

Chemistry

2 answers:

Rus_ich [418]3 years ago

7 0

Answer:

You can find the most reactive metals in the periodic table by looking for metals called alkali metals.

Explanation:

Alkali metals are chemical elements of group 1A of the periodic table with similar properties. The group consists of the following metals: lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs) and frâncio (Fr). They are low density metals, and moles. They are electropositive and highly reactive.

To know whether one metal is more reactive than another, it is important to remember the concept of electropositivity, that is, the ability of one atom to yield electrons to another atom during a bond. Thus, if one metal is more electropositive than another, it has a greater tendency to lose electrons in a bond and therefore tends to be more reactive.

faltersainse [42]3 years ago

5 0

The most reactive metals on the periodic table would have to be the alkali metals.
I hope this helps!
Have a nice day (:

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calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

3 years ago

4. The reaction of silver nitrate and potassium bromide yields silver bromide and potassium nitrate. If

Hatshy [7]

Answer:

1.) AgNO₃

2.) 0.563 moles AgBr

Explanation:

The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).

AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)

Molarity (M) = moles / liters

100 mL = 1 L

AgNO₃

45.0 mL / 100 = 45.0 L

1.25 M = ? moles / 0.450 L

? moles = 0.563 moles

KBr

75.0 mL / 100 = 0.750 L

0.800 M = ? moles / 0.750 L

? moles = 0.600 moles

In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.

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Zinaida [17]

washing machine, hopefully this helps

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If a drop of blood is 0.05 mL, how many drops of blood are in the blood collection tube that holds 2 mL?

Irina18 [472]

40 drops of blood in a tube that holds 2 mL

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3 years ago