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3 years ago

Chlorine has a atomic number of 17. It often forms an ion by gaining 1 electron. What would it's charge be?

Chemistry

1 answer:

kvasek [131]3 years ago

5 0

Answer:

The charge would be -1.

Explanation:

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The Mass of the mineral is 9.6 grams . The mineral is placed in a graduated cylinder containing 8.0 ml of water . The water rise

ryzh [129]

Answer:

1.2 g/ ml

Explanation:

The volume of the mineral = increase in volume of the water whuich is 16 - 8 = 8mls.

Therefore the mineral's density = 9.6 / 8

= 1.2 g/ ml answer

7 0

3 years ago

A chemical reaction that has the general formula of AB + CB + Als best classified as a

d1i1m1o1n [39]

Replacement I think, hope this helps ;)

Explanation:

4 0

3 years ago

A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm

VARVARA [1.3K]

<u>Answer:</u> The nuclear equations for the given process is written below.

<u>Explanation:</u>

The chemical equation for the bombardment of neutron to U-238 isotope follows:

$_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}$

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

The chemical equation for the first beta decay process of $_{92}^{239}\textrm{U}$ follows:

$_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta$

The chemical equation for the second beta decay process of $_{93}^{239}\textrm{Np}$ follows:

$_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta$

Hence, the nuclear equations for the given process is written above.

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

Three factors! that affect the solubility of a substance are pressure, the type of solvent, and volume.

lozanna [386]

Answer: yes even temperature too in rarely case

7 0

3 years ago