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3 years ago

I HAVE 6 MORE PAGES ON THIS HUUUUUUGE ASSIGNMENT!! WHO WANTS TO HELP?????????? i have alr done 22 pages!!

Chemistry

2 answers:

torisob [31]3 years ago

8 0

Answer:

it depends on the assignment but im down

Explanation:

Natalija [7]3 years ago

6 0

Answer:

what is it about I can prob help :)

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What is the only letter not appearing on the periodic table?.

ivolga24 [154]

Answer:

The answer is j

Explanation:

3 0

3 years ago

The equilibrium constant for the reaction

FinnZ [79.3K]

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

Explanation:

Moles of $Cl_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol$

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 :

$4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}$

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of $Cl_2$ at equilibrium is 0.023 M

7 0

3 years ago

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

3 years ago

How much heat is required to raise the temperature of 81.0 g of water from its melting point to its boiling point?

Dovator [93]

Answer:

Specific heat of water = 33.89 KJ

Explanation:

Given:

mass of water = 81 gram

Initial temperature = 0°C

Final temperature = 100°C

Specific heat of water = 4.184

Find:

Required heat Q

Computation:

Q = Mass x Specific heat of water x (Final temperature - Initial temperature)

Q = (81)(4.184)(100-0)

Q = 33,890.4

Specific heat of water = 33.89 KJ

6 0

3 years ago

26 Which process occurs in an operating voltaic cell?(1) Electrical energy is converted to chemical energy.

tia_tia [17]

Answer is: (2) Chemical energy is converted to electrical energy.

An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.

In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

8 0

3 years ago

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