K2CrO4 is added to a solution containing lead (II) and barium ions. If a precipitate is formed, what is it?

You start with 1 L of CO2 at standard temperature and pressure in a closed container. If you raise the temperature of the gas, t

pychu [463]

Answer:

Increase

Explanation:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

If the initial temperature and pressure is standard,

Pressure = 1 atm

Temperature = 273.15 K

then we increase the temperature to 400.0 K, The pressure will be,

1 atm / 273.15 K = P₂/400.0K

P₂ = 1 atm × 400.0 K / 273.15 K

P₂ = 400.0 atm. K /273.15 K

P₂ = 1.46 atm

Pressure is also increase from 1 atm to 1.46 atm.

8 0

2 years ago

A mixture of hydrogen (2.02 g) and chlorine (35.90 g) in a container at 300 K has a total gas pressure of 748 mm Hg. What is the

Llana [10]

The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.

<h3>How do we calculate the partial pressure of gas?</h3>

Partial pressure of particular gas will be calculated as:

p = nP, where

P = total pressure = 748 mmHg
n is the mole fraction which can be calculated as:
n = moles of gas / total moles of gas

Moles will be calculated as:

n = W/M, where
W = given mass
M = molar mass

Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole

Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole

Mole fraction of hydrogen = 1 / (1+0.5) = 0.6

Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm

Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.

To know more about partial pressure, visit the below link:
brainly.com/question/15302032

#SPJ1

3 0

2 years ago

Which equation shows how to solve for enthalpy of solution based on the information in tables?

Tatiana [17]

Answer: B. triH sol Mgl2= -triHlat+ triHhydr Mg^2+ 2triHhydr^l-

Explanation:

Just did it and it was right

8 0

3 years ago

At 9°C a gas has a volume of 6.17 L. What is its volume when the gas is at standard temperature?

Alex17521 [72]

Answer:

V₂ = 5.97 L

Explanation:

Given data:

Initial temperature = 9°C (9+273 = 282 K)

Initial volume of gas = 6.17 L

Final volume of gas = ?

Final temperature = standard = 273 K

Solution:

Formula:

The Charles Law will be apply to solve the given problem.

According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 6.17 L × 273K / 282 k

V₂ = 1684.41 L.K / 282 K

V₂ = 5.97 L

5 0

3 years ago

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): 2H2O2

zalisa [80]

Answer:

23.0733 L

Explanation:

The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

$Mass=\frac {50}{100}\times 125\ g$

Mass = 62.5 g

Molar mass of $H_2O_2$ = 34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{62.5\ g}{34\ g/mol}$

$moles= 1.8382\ mol$

Consider the given reaction as:

$2H_2O_2_{(aq)}\rightarrow2H_2O_{(l)}+O_2_{(g)}$

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.

Also,

1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.

So,

1.8382 moles of hydrogen peroxide decomposes to give $\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr
The conversion of P(torr) to P(atm) is shown below:
[tex]P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 746 / 760 atm = 0.9816 atm

Temperature = 27 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (27 + 273.15) K = 300.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K

8 0

3 years ago