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KatRina [158]
3 years ago
7

When Bruce Banner tried saving his friend from a gamma bomb blast, he was hit with a large dose of gamma radiation. Following th

e blast, he turned into The Incredible Hulk. Gamma radiation has a wavelength of 1.00 x 10-12 m. Calculate the frequency of the waves.
Chemistry
1 answer:
ivanzaharov [21]3 years ago
7 0

Answer:

22

Explanation:

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What type of elements are generally involved in covalent bonding?
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<span>In organic chemistry, covalent bonding is most often associated with carbon compounds, which are known as organic chemicals. Hydrogen is also involved most of the time, as well as oxygen. Other elements can also be involved, but less frequently.</span>
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What shape is the s sublevel? *<br> sphere<br> freaky<br> peanut<br> daisy
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Answer:

Sphere

Explanation:

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14.28 explain how you would distinguish between each pair of compounds using high- resolution mass spectrometry
vagabundo [1.1K]

The distinguish between each pair of compounds using high- resolution mass spectrometry by the exact mass rather than nominal mass are utilizes to measure the compound.

The mass spectrometry is involves the following steps :

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Mass spectrometry is the analytical method useful for the calculating the mass to charge ratio ( m / z ). the mass spectrometry is based on the newton's second law and the momentum.

Thus, the mass spectroscopy is method to measure the molecular mass of the compound and indirectly helps examine the isotopes and based on the newton's second law .

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7 0
1 year ago
Can someone help me please
MArishka [77]

Answer:

Group 4A (or IVA) of the periodic table includes the nonmetal carbon (C), the metalloids silicon (Si) and germanium (Ge), the metals tin (Sn) and lead (Pb), and the yet-unnamed artificially-produced element ununquadium (Uuq).

The Group 4A elements have four valence electrons in their highest-energy orbitals (ns2np2). Carbon and silicon can form ionic compounds by gaining four electrons, forming the carbide anion (C4-) and silicide anion (Si4-), but they more frequently form compounds through covalent bonding. Tin and lead can lose either their outermost p electrons to form 2+ charges (Sn2+, the stannous ion, and Pb2+, the plumbous ion) or their outermost s and p electrons to form 4+ charges (Sn4+, the stannic ion, and Pb4+, the plumbic ion).

Carbon (C, Z=6).

Carbon is most familiar as a black solid is graphite, coal, and charcoal, or as the hard, crystalline diamond form. The name is derived from the Latin word for charcoal, carbo. It is found in the Earth's crust at a concentration of 480 ppm, making it the 15th most abundant element. It is found in form of calcium carbonate, CaCO3, in minerals such as limestone, marble, and dolomite (a mixture of calcium and

Explanation:

<em><u>T</u></em><em><u>H</u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>A</u></em><em><u>L</u></em><em><u>L</u></em><em><u> </u></em><em><u>I</u></em><em><u> </u></em><em><u>K</u></em><em><u>N</u></em><em><u>O</u></em><em><u>W</u></em>

<u>E</u><u>N</u><u>J</u><u>O</u><u>Y</u><u> </u><u>THE</u><em><u> </u></em><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em>

3 0
3 years ago
A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate
Vika [28.1K]

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

0.030 cm³ × ? = x m³

You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

1 cm = 10⁻² m

If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.

So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.

We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}

7 0
3 years ago
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