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4 years ago

What would the molecule CH₄ be classified as?

Chemistry

1 answer:

larisa [96]4 years ago

6 0

Answer:

Alkane

Explanation:

Definition of Alkane "any of the series of saturated hydrocarbons including methane, ethane, propane, and higher members. (google dictionary)"

CH4 is methane.

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The Sun is the primary source of energy that influences any planet's temperature, including Earth. The amount of energy received from the Sun is called insolation; the ratio reflected is called the albedo.

Explanation:

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4 years ago

What is Rutherford atomic modle

Ksju [112]

Answer:

Rutherford's atomic model explained how the electrons surrounded the nucleus of protons and neutrons. His model showed how J. J. Thomson's Plum Pudding model was incorrect.

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3 years ago

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A 72 g sample contains how many moles of water? (include appropriate sig figs in your answer)

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Answer:

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3 years ago

Chlorine has two natural isotopes, chlorine-35 and chlorine-37.

Alex Ar [27]

Answer:

The relative atomic mass of chlorine depends on the ratio between the abundance of these two naturally-occurring isotopes.

Explanation:

The relative atomic mass of an element is a weighted average of the atomic mass of its naturally-occurring isotopes. The relative abundance of each isotope gives its weight in this weighted average.

For these two naturally-occurring isotopes of chlorine:

The relative atomic mass of $^{35}{\rm Cl}$ is approximately $34.969$ daltons. The relative abundance of this isotope in nature is approximately $0.758$ .
The relative atomic mass $^{37}{\rm Cl}$ is approximately $36.966$ daltons. The relative abundance of this isotope in nature is approximately $0.242$ .

$\begin{array}{|c|c|c|}\cline{1-3} \text{Isotope} & \text{Atomic Mass} & \text{Relative Abundance}\\ \cline{1-3} ^{35}{\rm Cl} & \approx 34.968\; \rm Da} & \approx 0.758 \\ \cline{1-3} ^{37}{\rm Cl} & \approx 36.966\; \rm Da & \approx 0.242 \\ \cline{1-3}\end{array}$ .

$\begin{aligned}&\text{relative atomic mass of Cl} \\ &= \text{atomic mass of $^{35}{\rm Cl}$} \times \text{relative abundance of $^{35}{\rm Cl}$} \\&\quad + \text{atomic mass of $^{37}{\rm Cl}$} \times \text{relative abundance of $^{37}{\rm Cl}$} \\ &\approx 34.968\; \rm Da \times 0.758 + 36.966\; \rm Da \times 0.242 \\ &\approx 35.45\; \rm Da \end{aligned}$ .

The relative abundance of $^{35}{\rm Cl}$ is much higher than that of $^{37}{\rm Cl}$ . Consequently, the relative atomic mass of the element $\rm Cl$ is closer to the atomic mass of $^{35}{\rm Cl}\!$ than that of $^{37}{\rm Cl}\!$ .

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3 years ago

Given that a particular photon has a frequency of 2.2 x 10'7 Hz, calculate the photon's energy. Submit your answer in

BlackZzzverrR [31]

Answer:

E = 15×10⁻²⁹ J

Explanation:

Given data:

Frequency of photon = 2.2× 10⁷ Hz

Energy of photon = ?

Solution:

Formula:

E = h.f

h = 6.63×10⁻³⁴ Js

by putting values,

E = 6.63×10⁻³⁴ Js × 2.2× 10⁷ s⁻¹

E = 14.586 ×10⁻²⁹ J

E = 15×10⁻²⁹ J

The energy of photon is 15×10⁻²⁹ J.

5 0

3 years ago