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Alex787 [66]
3 years ago
7

Alkanes are hydrocarbons that contain _____.

Chemistry
1 answer:
abruzzese [7]3 years ago
6 0
I would have to say D) carbon-to-hydrogen ratio
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How do you calculate the number of valence electrons in an atom?
gavmur [86]
To find<span> the </span>valence electrons in an atom<span>, identify what group the element is in. An element in group 1A has 1 </span>valence electron<span>. For example, Li is in group 1A, so that means it has one </span>valence electron. If the element is in group 2A, then it has two valence electrons<span>.</span>
5 0
3 years ago
What is properties of matter​
alexandr1967 [171]

Explanation:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils

melting point (intensive): the temperature at which a substance 

7 0
3 years ago
Read 2 more answers
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
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What did organizing information on the periodic table reveal?
Natalka [10]

Answer:

The groupings of elements in the periodic table reveal various trends, present in groups such as the alkali metals and noble gases, or the lanthanides and actinides. The periodic table can be used to find information on elements or predict the properties of as-of-yet undiscovered elements.

Explanation:

4 0
3 years ago
Read 2 more answers
A steel beam that is 7.00 m long weighs 326 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
GrogVix [38]

Answer:

The answer to the question is

Suki is 1.12 m close to the end of the steel beam before the beam begins to tip

Explanation:

To solve the question we list out the known variables first

Length of steel beam, L = 7.00 m

Weight  of steal beam W = 326 N

Distance between supports = 3.00 m

Weight of Suki = 555 N

By moments theory, the sum of moments about a point = 0

and assuming the steal beam is uniform, the weight of the beam will act at the center of the beam

As described the supports are 2.00 m from each end of the beam thus

taking moments about one of the support with Suki between  the end of the beam and the support we have

Distance from center of beam to support = 1.5 m

Distance of Suki to the nearest support = x m

therefore 555 × x + 326 × 1.5 = 0

or x = -0.881 m on the other side of the support

Therefore Suki is (2 - 0.881) m or 1.12 m close to the end before the beam begins to tip

7 0
3 years ago
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