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3 years ago

Which of the following is true regarding the transition elements?

Chemistry

1 answer:

e-lub [12.9K]3 years ago

8 0

Pls complete your question

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The chemical symbol for a calcium ion is ca2 . what does the 2 symbolize

Blababa [14]

2 stands for valency of ca ion

its +2

that it takes two electrons to complete its shell

8 0

3 years ago

What is the mass of oxygen that can be produced from 2.79 moles of lead(ll) nitrate

denis23 [38]

1.38 moles of oxygen

Explanation:

Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

The mole ration of Lead (II) nitrate to oxygen is 2: 1

Therefore 2.76 moles of Lead (II) nitrate will lead to production of? moles of oxygen;

2: 1

2.76: x

Cross-multiply;

2x = 2.76 * 1

x = 2.76 / 2

x = 1.38

8 0

3 years ago

A sample of water at 20 degrees c contains what bonds

vladimir2022 [97]

Covalent and hydrogen bonds

3 0

3 years ago

When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200

GaryK [48]

Answer:

$2.023 m^3$ is the total displacement (volume) of the submarine.

Explanation:

Mass of water carried by submarine at 1000 ft depth = m = 2100 kg

The density of seawater at 1000 ft depth = d = $1033 kg/m^3$

Volume of the water displaced = V= ?

Total displacement of the submarine = Volume of the water displaced = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3$

$2.023 m^3$ is the total displacement (volume) of the submarine.

6 0

3 years ago

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