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aalyn [17]
2 years ago
9

An Asteroid is smaller than a planet but larger than a meteoroid? True or false pls no files or links they DO NOT WORK!!!!

Chemistry
1 answer:
velikii [3]2 years ago
4 0

Answer:

true astroids are in space and are bigger than meteorites but smaller than planets, meteorites which are asteroids after landing on earth and losing most of its mass on its way

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Help! please! hurry! I will mark brainliest if you get it correct!! NO SPAM!!!!!!
Troyanec [42]

Answer: 5.747 * 10^14 Hz

Explanation:

Convert 522nm to m = 522 * 10^-9 m (since 1nm=10^-9m)

If c = wavelength * frequency, where c is the speed of light (3.0 * 10^8 m/s), then you can manipulate the equation to solve for frequency (f).

f = c / wavelength

Plug in the given numbers:

f = (3.0 * 10^8) / (10^-9)

f = 5.747 * 10^14 Hz

4 0
2 years ago
Read 2 more answers
What type of radiation is emitted when chromium-51 decays into manganese-51? Show the nuclear equation that leads you to this an
Anastasy [175]
It's a beta decay
24Cr51 --> 25Mn51 + -1eo
4 0
3 years ago
Read 2 more answers
Which spheres are part of the earth system
zubka84 [21]

Answer:

Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" (land), "hydrosphere" (water), "biosphere" (living things), and "atmosphere" (air).

Explanation:

8 0
3 years ago
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of
Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH}  =0.1900gO

4. Compute the moles of oxygen by using its molar mass:

n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1

6. Search for the closest whole number (in this case multiply by 2):

C_3H_6O_2

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

M=12*3+1*6+16*2=74g/mol

Which is about three times in the molecular formula, for that reason, the actual formula is:

C_9H_{18}O_6

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0
3 years ago
The Oxidation number of iron in the Complexion [Fe(CN) 6]³^- is what?​
LUCKY_DIMON [66]

+3

Hence, the magnetic behaviour of the complex is paramagnetic. The oxidation number of the central metal atom: The oxidation number of the metal iron is +3.

4 0
2 years ago
Read 2 more answers
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