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3 years ago

6

Chemistry

1 answer:

tino4ka555 [31]3 years ago

5 0

Answer : The correct option is, electron.

Explanation :

As we know that,

An atom is the smallest unit of a matter that consist of three subatomic particles which are electrons, protons and neutrons.

The protons and the neutrons are located inside the nucleus or the center of the nucleus where the mass of the an atom is concentrated.

The electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and the neutrons are neutral that means it has no charge.

According to the question, the electron subatomic particle found revolving around the nucleus.

Hence, the correct option is electron.

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A solution containing a nonvolatile solute will have

svetoff [14.1K]

It is nonvolatile solute

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3 years ago

You wish to make a 0.299 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid

Art [367]

Answer:

$V_1=2.49mL$

Explanation:

Hello,

In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:

$n_{HI}=n_{HI}$

One could apply the following equality in terms of molarity:

$V_1M_1=V_2M_2$

Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:

$V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL$

Best regards.

5 0

3 years ago

Helpppppppppppppp!!!!!!!

OlgaM077 [116]

Something on a man i think it’s crocs

3 0

3 years ago

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

HELP ASAP!!!!!!!

Mrrafil [7]

"One has more oxygen atoms than the other" best distinguishes carbon dioxide from carbon monoxide

<u>Explanation</u>:

In both carbon monoxide and carbon dioxide, there is the presence of carbon and oxygen. However,they vary depending on the number of oxygen atoms.In carbon monoxide, there is the presence of one only carbon and oxygen atom. Thus carbon monoxide is chemically represented as CO. Only about 0.2%ppm of carbon monoxide present in air. carbon monoxide is toxic and causes death in human when not consumed is correct proportion.

Whereas in carbon dioxide, only one carbon atom,and two oxygen atom and hence chemically represented as co2.

3 0

3 years ago