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3 years ago

12. How much mass is in a 3.25-mole sample of NH 4 OH? A. 10.8 g B. 34.0 g C. 35.1 g D. 114 g

Chemistry

1 answer:

Galina-37 [17]3 years ago

8 0

Answer:

D. 114 g

Explanation:

NH4OH molecular weight. Molar mass of NH4OH = 35.0458 g/mol This compound is also known as Ammonium Hydroxide.
Convert grams NH4OH to moles or moles NH4OH to grams. Molecular weight calculation: 14.0067 + 1.00794*4 + 15.9994 + 1.00794.

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Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric che

Helga [31]

Answer:

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

Explanation:

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

Initially

3.0 atm 0

At equilibrium

(3.0-2p) p

Equilibrium partial pressure of $NO_2=2.1atm=3.0-2p$

p = 0.45 atm

The value of equilibrium constant wil be given by :

$K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}=\frac{p}{(3.0-2p)^2}$

$K_p=\frac{0.45}{(2.1)^2}=0.10$

After addition of 1.5 atm of nitrogen dioxide gas equilibrium reestablishes it self :

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

After adding 1.5 atm of $NO_2$ :

(2.1+1.5) atm 0.45 atm

At second equilibrium:'

(3.6-2P) (0.45+P)

The expression of equilibrium can be written as:

$K_p=\frac{p'_{N_2O_4}}{(p'_{NO_2})^2}$

$0.10=\frac{(0.45+P)}{(3.6-2P)^2}$

Solving for P:

P = 0.37 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time:

= (0.45+P) atm = (0.45 + 0.37 )atm = 0.82 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

5 0

3 years ago

In the laboratory, hydrogen gas is usually made by the following reaction: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) How many liters

IrinaK [193]

<u>Answer:</u> The volume of hydrogen gas collected over water is 2.13 L

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

$\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol$

For the given chemical reaction:

$Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)$

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = $\frac{1}[1}\times 0.0851=0.0851mol$ of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

Putting values in above equation, we get:

$733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L$

Hence, the volume of hydrogen gas collected over water is 2.13 L

5 0

3 years ago

What is developmental delay? why is it so important to respond promptly when one is suspected?

Vadim26 [7]

Developmental delays is when your child dose not reach there developmental milestones at the expected times...It is important to respond promptly so that the child can get the help they need to help them get back on the right track

8 0

3 years ago

From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l)

disa [49]

Answer:

618 J/Kmol

T > 1.36 x 10³ K

Explanation:

The balanced reaction of interest is:

SO₃ (g) + SCl₂ (l) ⇒ SOCl₂ (l) + SO₂ (g)

with the data:

ΔHºf (kJ/mol) -396 -50.0 -245.6 -296.8

Sº(J/mol·K) 256.7 184 ? -248.1

ΔGº= -75.2 kJ

We know, we can find the standard change inGibb´s free energy from the equation:

ΔGºrxn = ΔHºrxn - TΔSºrxn

So we can calculate ΔHºrxn = ∑ ΔHºf prod - ΔHºreact, and substitute into this equation to solve Sº SOCl₂.

ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ

Similarly for ΔSºrxn

ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K

= -0.689 kJ /K -+ Sº SOCl₂

Plugging the values for the expression for ΔGºrxn:

-75.2 kJ = -96.4 kJ - 298 K x ( -0.689 kJ /K + Sº SCl₂ )

-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂

-184 kJ = -298 K x Sº SCl₂

0.618 kJ/molK = Sº SCl₂

= 0.618 kJ/K x 1000 J = 618 J/Kmol

For the second part we will still be using the Gibb´s free energy change equation as above , but now we will solve for T when the reaction becomes non-spontaneous.

For the reaction to become non-spontaneous ΔGº is positive, and this happens when the term TΔSº becomes greater tha ΔHº:

ΔGºrxn = ΔHºrxn - TΔSºrxn

0 = ΔHºrxn - TΔSºrxn ⇒ TΔSºrxn = ΔHºrxn

T= ΔHºrxn / ΔSºrxn

ΔSºrxn = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol

( using the value the value just calculated from above )

T = - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K

For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.

4 0

4 years ago

The chain of events in a story is known as the _________

Ratling [72]

The term you are referring to is the plot

5 0

3 years ago