Ask question

Ask question

All categories

3 years ago

5

Choose the correct statement about melting points. * Melting point can tell us if we have a mixture of compounds (for example, c

ompound of interest impurity). Melting point can tell us the identity of the components of a mixture. If we have a sample with a lowered melting range as compared to our standards, we can do mixed melting point determinations, combining our sample with each standard, to determine the identity of our sample. All of these are true. None of these are true.

1 answer:

goldfiish [28.3K]3 years ago

4 0

Answer: All of these statements are true

Explanation:

Melting point help us to determine if a mixture is pure or has impurities by the virtues of it melting range..

You might be interested in

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

How does an electron emit light?

Semenov [28]

Answer:

by moving between defined energy levels

6 0

3 years ago

PLZZZZZZZ HELPPPP MEEEE ASAP CHEMISTRY QUESTION DOWN BELOW How can you identify a reversible reaction from a chemical equation?

zalisa [80]

it's the third one The chemical equation will have two arrows pointing in different directions. !

7 0

3 years ago

Read 2 more answers

Which of the following elements would you expect to be extremely explosive in the presence of water?

Natali5045456 [20]

The answer is a. Sodium

6 0

3 years ago

Read 2 more answers

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

nikdorinn [45]

Answer : The volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

Explanation :

The balanced chemical reaction will be:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

First we have to calculate the moles of Al.

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

Molar mass of Al = 27 g/mole

$\text{Moles of }Al=\frac{55g}{27g/mol}=2mole$

Now we have to calculate the moles of $O_2$ .

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of $O_2$

So, 2 mole of Al react with $\frac{3}{4}\times 2=0.75\times 2$ moles of $O_2$

Now we have to calculate the volume of $O_2$ consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of $O_2$ occupies 22.4 liter volume of $O_2$ gas

So, $0.75\times 2$ mole of $O_2$ occupies $0.75\times 2\times 22.4$ liter volume of $O_2$ gas

Therefore, the volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

3 0

3 years ago