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4.00 dm of oxygen at a pressure of 2 atm and 1.00 dm3 nitrogen at a pressure of 1 atm are introduced into a 2.00 dm3 vessel.

podryga [215]

Answer:

A

Explanation:

5 0

3 years ago

Which of the following chemical formulas represents a total of 12 atoms of oxygen?

9966 [12]

Answer:

3Sn(s04)2

Explanation:bc it cant be anything else

4 0

3 years ago

Read 2 more answers

How many protons and electrons are in Hg+

Gala2k [10]

Answer:

There are 80 protons and 80 electrons in Hg+ (Mercury)

Explanation:

Name Mercury

Symbol Hg

Atomic Number 80

Atomic Mass 200.59 atomic mass units

Number of Protons 80

Number of Neutrons 121

Number of Electrons 80

Melting Point -38.87° C

Boiling Point 356.58° C

Density 13.456 grams per cubic centimeter

Normal Phase Liquid

Family Transitions Metals

Period Number 6

8 0

3 years ago

What is the change in the freezing point of water when 35.0 g of sucrose is

frez [133]

From the calculations, we can see that, the change in the freezing point is -0.634°C.

<h3>What is freezing point?</h3>

The term freezing point refers to the temperature at which a liquid is changed to solid.

Given that;

ΔT = K m i

Number of moles sucrose = 35.0 g/ 342.30 g/mol = 0.1 moles

molality = 0.1 moles/ 300.0 * 10^-3 Kg

= 0.33 m

Thus;

ΔT = -1.86°C/mol * 0.33 m * 1

= -0.634°C

Learn more about freezing point:brainly.com/question/3121416

#SPJ1

3 0

2 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