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2 years ago

15

A + B + C → D + E

2 answers:

brilliants [131]2 years ago

8 0

I believe it is B... because all balance equation are supposed to follow the law of conservation of mass

Alexeev081 [22]2 years ago

5 0

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

For the given chemical reaction:

$A+B+C\rightarrow D+E$

Total mass on reactant side = Total mass on product side

Hence, the correct answer is Option B.

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The motion of atoms and molecules that most affects temperature is

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3 years ago

The total volume of seawater is 1.5 x 1021L .Assume that seawater contains 3.1 percent sodium chloride by mass and that its dens

son4ous [18]

Answer:

Mass in kg = 4.7*10^19 kg

Mass in tons = 5.2*10^16 tons

Explanation:

<u>Given:</u>

Total volume of sea water = 1.5*10^21 L

Mass % NaCl in seawater = 3.1%

Density of seawater = 1.03 g/ml

<u>To determine:</u>

Total mass of NaCl in kg and in tons

<u>Calculation:</u>

Unit conversion:

1 L = 1000 ml

The volume of seawater in ml is:

$=\frac{1.5*10^{21}L*1000ml }{1L} =1.5*10^{24} ml$

$Mass\ seawater = Density*volume = 1.03g/ml*1.5*10^{24} ml=1.5*10^{24}g$

$Mass\ NaCl\ in\ seawater = \frac{3.1}{100}*1.5*10^{24} g=4.7*10^{22} g$

To convert mass from g to Kg:

1000 g = 1 kg

$Mass\ seawater(kg) = \frac{4.7*10^{22}g*1kg }{1000g} =4.7*10^{19} kg$

To convert mass from g to tons:

1 ton = 9.072*10^6 g

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5 0

3 years ago

Which of the following best describes an example of applied chemistry?

hram777 [196]

Answer:

Explanation:

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2 years ago

The equilibrium constant for the reaction

FinnZ [79.3K]

Answer: The concentrations of $Cl_2$ at equilibrium is 0.023 M

Explanation:

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Volume of solution = 1 L

Initial concentration of $Cl_2$ = $\frac{0.14mol}{1L}=0.14M$

The given balanced equilibrium reaction is,

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial conc. 0.14 M 0 M 0M

At eqm. conc. (0.14-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}$

Now put all the given values in this expression, we get :

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By solving the term 'x', we get :

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3 years ago

How is kinetic energy related to the Kelvin scale?

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3 years ago