Please help me I’m desperate plz help and fast

Ill give u brainliest if u help asap

UkoKoshka [18]

Radio waves, radios

microwaves, microwaves

Infrared, tv remotes

5 0

3 years ago

Calculate the oxidation numbers of the elements in this equation 2 Al2O3 → 4 Al + 3 O2

riadik2000 [5.3K]

Explanation:

Answer

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The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0

2× ( Oxidation number of Al) +3(−2)=0

2× ( oxidation number of Al) +6

∴ Oxidation number of Al =+3

5 0

3 years ago

What process is used to release energy in nuclear power plants?

lukranit [14]

Nuclear fission is used enable to release energy in power plants. The constant collision of particles within the reactor, create most of the plants energy.

6 0

3 years ago

Read 2 more answers

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t

topjm [15]

Answer:

$\boxed{3}$

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

$\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )$

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

$\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}$

Data:

$n_{f} = 2$

λ = 657 nm

Calculation:

$\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}$

7 0

3 years ago

What is the volume of 14.0g of nitrogen gas at STP?

lozanna [386]

Answer:

The volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

1) Convert 14.0 g of nitrogen gas to number of moles:

n = mass in grams / molar mass

Atomic mass of nitrogen: 14.0 g/mol

Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol

n = 14.0 g / 28.0 g/mol = 0.500 mol

2) Set a proportion to calculate the volume of nitrogen gas:

22.4 liter / mol = x / 0.500 mol

Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

Conclusion: the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

6 0

3 years ago