All categories

3 years ago

2.50 x 10-14 \div÷ 1.25 x 10-12

Chemistry

1 answer:

juin [17]3 years ago

5 0

Answer:

Answer is 22.

Explanation:

=2.50×10-14÷1.25×10-12

=25-14÷12.5-12

=22

You might be interested in

example best shows that the chemistry of water is helpful to plants? Water’s polarity produces a high density, which allows wate

Gennadij [26K]

The right answer for the question that is being asked and shown above is that: "Water’s polarity produces a high density, which allows water to move to the leaves." Example best shows that the chemistry of water is helpful to plants is that <span>Water’s polarity produces a high density, which allows water to move to the leaves.</span>

8 0

3 years ago

Read 2 more answers

NEEED HELP PLEASE IM VERY STRESSED

Arte-miy333 [17]

Answer:

C) It moves faster and collects more data

Explanation:

5 0

2 years ago

What is the name of this hydrocarbon? 2–methylpentyne 2–dimethylpentene 2,3–dimethylpentane 3,2–dimethylpentane

Phantasy [73]

<span>on plato it's 2,3–dimethylpentane

</span>

6 0

2 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

2 years ago

What is the element with 2 protons on the

Paul [167]

Answer:

Helium

Explanation:

7 0

2 years ago