In this reaction which substance behaves as the oxidizing agent pb

VashaNatasha [74]

The Inclosure Acts, which use an old or formal spelling of the word now usually spelt "enclosure", cover enclosure of open fields and common land in England and Wales, creating legal property rights to land previously held in common.

4 0

3 years ago

Suppose that 33.3 J of heat is added to an ideal gas. The gas expands at a constant pressure of 1.45 104 Pa while changing its v

levacccp [35]

Answer:

ΔU = 25.8 J

Explanation:

The gas absorbs 33.3 J of heat, that is, Q = 33.3 J.

The work (W) of expansion can be calculated using the following expression:

W = -P. ΔV

where,

P is the external pressure

ΔV is the change in volume

W = -1.45 × 10⁴ N . m⁻² × (8.40 × 10⁻⁴ m³ - 3.24 × 10⁻⁴ m³) = -7.48 J

The change in the internal energy (ΔU) is:

ΔU = Q + W

ΔU = 33.3 J + (-7.48 J) = 25.8 J

5 0

3 years ago

A serving of Cheez-Its releases 1.30 x 10^4 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy we

kondor19780726 [428]

Answer:

The final temperature is:- 7428571463.57 °C

Explanation:

The expression for the calculation of heat is shown below as:-

$Q=m\times C\times \Delta T$

Where,

$Q$ is the heat absorbed/released

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)

Specific heat of water = 4.18 J/g°C

Initial temperature = 35 °C

Final temperature = x °C

$\Delta T=(x-35)\ ^0C/tex]
Q = [tex]1.3\times 10^4$ kcal

Also, 1 kcal = 4.18 kJ = $4.18\times 10^3$ J

So, Q = $1.3\times 10^4\times 4.18\times 10^3$ J = 54340000 J

So,

$54340000=0.00175\times 4.18\times (x-35)$

$0.00175\times \:4.18\left(x-35\right)=54340000$

$x-35=\frac{54340000}{0.007315}$

$x=7428571463.57$

Thus, the final temperature is:- 7428571463.57 °C

3 0

3 years ago

Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of

Oxana [17]

<span>12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Alâ†’3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g</span>

7 0

3 years ago

If Oxygen were to gain 2 electrons, the new electron configuration for the oxide ion

Naddika [18.5K]

It would be most similar to neon. it wouldn’t be sulfur because that’s in the same group as oxygen and has the same number of electrons. and carbon has less than that so the only one that makes sense is neon

5 0

3 years ago