What is the oxidation number for O in H2O2 ?

Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from

aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

Initial volume as 2.0 L
Initial temperature as 60.0°C
New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where, $V_1 and V_2$ are initial and new volume respectively, while $T_1 and T_2$ are initial and new temperatures respectively;

$T_1= 333 K$

$T_2=303K$

$V_1 =2.0L$

Rearranging the formula;

$V_2=\frac{V_1T_2}{T_1}$

$= \frac{(2.0L)(303K)}{333K} \\=1.820 L$

Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0

2 years ago

Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 3.0 moles of N2 react with

Alex787 [66]

Answer:

Work done in Joules = -14865.432J

ΔU = -262.94KJ

Explanation:

The steps are as shown in the attachment

6 0

3 years ago

Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

GarryVolchara [31]

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

6 0

3 years ago

What is Avogadro's number, and why is it useful? (3 points)

SashulF [63]

Answer:

6.022 x 10²³; it is a conversion factor between moles and number of particles

Explanation:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole of hydrogen = 6.022 × 10²³ atoms of hydrogen

238 g of uranium = 1 mole of uranium = 6.022 × 10²³ atoms of uranium

By taking ions:

62 g of NO⁻₃ = 1 mole of NO⁻₃ = 6.022 × 10²³ ions of NO⁻₃

96 g of SO₄²⁻ = 1 mole of SO₄²⁻ = 6.022 × 10²³ ions of SO₄²⁻

4 0

3 years ago

After substantial heating, 6.25 g of iron produced 18.00 g of a compound with chlorine. The empirical formula is:

babunello [35]

Answer:

Option A. FeCl3

Explanation:

The following data were obtained from the question:

Mass of iron (Fe) = 6.25g

Mass of the compound formed = 18g

From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow

Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.

Mass of chlorine (Cl) = 18 – 6.25

Mass of chlorine (Cl) = 11.75g

The compound therefore contains:

Iron (Fe) = 6.25g

Chlorine (Cl) = 11.75g

The empirical formula for the compound can be obtained by doing the following:

Step 1:

Divide by their molar mass

Fe = 6.25/56 = 0.112

Cl = 11.75/35.5 = 0.331

Step 2:

Divide by the smallest

Fe = 0.112/0.112 = 1

Cl = 0.331/0.112 = 3

The empirical formula for the compound is FeCl3

3 0

3 years ago