What does inertia depend on??

Small quantites of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc.

ira [324]

<u>Answer:</u> The mass or zinc reacted is 0.624 grams.

<u>Explanation:</u>

We are given:

Total pressure = 1.032 atm

Vapor pressure of water = 32 torr = 0.042 atm (Conversion factor: 1 atm = 760 torr)

To calculate partial pressure of hydrogen gas, we use the equation:

$p_{H_2}=p_T-p_{H_2O}\\\\p_{H_2}=1.032-0.042=0.99atm$

To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of hydrogen gas = 0.99 atm

V = Volume of hydrogen gas = 240. mL = 0.240 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of hydrogen gas = $30^oC=[30+273]K=303K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$0.99atm\times 0.240L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303K\\n=\frac{0.99\times 0.240}{0.0821\times 303}=9.55\times 10^{-3}mol$

The chemical equation for the reaction of zinc and hydrochloric acid follows:

$Zn+2HCl\rightarrow ZnCl_2+H_2$

By Stoichiometry of the reaction:

1 mole of hydrogen gas is produced from 1 mole of zinc metal

So, $9.55\times 10^{-3}mol$ of hydrogen gas is produced from = $\frac{1}{1}\times 9.55\times 10^{-3}=9.55\times 10^{-3}mol$ of zinc metal

To calculate the mass of zinc metal, we use the equation:

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

Molar mass of zinc = 65.38 g/mol

Moles of zinc = $9.55\times 10^{-3}$ moles

Putting values in above equation, we get:

$9.55\times 10^{-3}mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=(9.55\times 10^{-3}mol\times 65.38g/mol)=0.624g$

Hence, the mass or zinc reacted is 0.624 grams.

4 0

3 years ago

1) Alarge doud of dust and gas spread out in a large area is

alexdok [17]

Answer:

D. nebula

Explanation:

Hope this helps

<3

8 0

2 years ago

You observe mothballs disappearing in cabinets. What do you think is the reason for this? Do all substances behave like mothball

cricket20 [7]

Answer: Mothballs have weak intermolecular forces.

No all substances do not behave like mothballs at normal conditions. Example: benzene , chloroform

Explanation:

Sublimation is a process of converting a substance from solid state to gaseous state without the formation of liquid at constant temperature.

A substance which undergoes sublimation is called as sublimating substance.

As mothballs is made of napthalene which has weak inter molecular forces of attraction between its molecules, it directly sublimes into gaseous state without leaving any residue and is called as a sublimating substance.

Not all substances behave like mothballs at normal conditions. Example: benzene , chloroform

8 0

3 years ago

1. If we used 0.0100 moles of K2CO3, how many moles of SrCO3 can be expected to form?

faltersainse [42]

Answer:

0.01 moles of SrCO₃

Explanation:

In this excersise we need to propose the reaction:

K₂CO₃ + Sr(NO₃)₂ → 2KNO₃ + SrCO₃

As we only have data about the potassium carbonate we assume the strontium nitrite as the excess reactant.

1 mol of K₂CO₃ react to 1 mol of Sr(NO₃)₂ in order to produce 2 moles of potassium nitrite and 1 mol of strontium carbonate.

Ratio is 1:1. In conclussion,

0.01 mol of K₂CO₃ must produce 0.01 moles of SrCO₃

3 0

3 years ago

A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut

leonid [27]

Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

=

x(t)

200

.

We model this problem as

dx

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

·

20 Lnit.

100 Lsol.

=

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

·

x(t) Lnit.

200 Lsol.

=

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dx

dt = 1.2 − 0.03x,

or

dx

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

= Ce−0.03t +

1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

3 0

3 years ago