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3 years ago

Which answer best explains the relationship between size and mass of the sun?

1 answer:

3 0

The mean radius of the sun is 432,450 miles (696,000 kilometers), which makes its diameter about 864,938 miles (1.392 million km). You could line up 109 Earths across the face of the sun. The sun's circumference is about 2,713,406 miles (4,366,813 km). The total volume of the sun is 1.4 x 1027 cubic meters.

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Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda

bezimeni [28]

<u>Answer:</u> The average atomic mass of X is 28.09 amu

<u>Explanation:</u>

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:

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

<u>For isotope 1:</u>

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

<u>For isotope 2:</u>

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

<u>For isotope 3:</u>

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]$

$\text{Average atomic mass of X}=28.09amu$

Hence, the average atomic mass of X is 28.09 amu

4 0

3 years ago

Matter and energy are

Wewaii [24]

When energy transforms into mass, the amount of energy does not remain the same. When mass transforms into energy, the amount of energy also does not remain the same. However, the amount of matter and energy remains the same. ... You would weigh much less on the Moon because it is only about one-sixth the mass of Earth. So the answer is D

3 0

2 years ago

Why is concentrated sulphuric acid is weak acid

frez [133]

Answer:

Sulfuric acid is a strong acid in any state

Explanation:

You may be thinking if the definition that a weak acid produces relatively few ions in aqueous solution,

If you have 100 % sulfuric acid, there is no water, so the definition does not apply.

Commercial sulfuric acid contains about 2 % water. The sulfuric acid in that small amount of water consists of about 99 % HSO₄⁻ and 1 % SO₄²⁻. The acid is almost completely ionized.

Sulfuric acid is a strong acid.

7 0

3 years ago

if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?

vlada-n [284]

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 2 moles
H₂SO₄: 3 moles
Al₂(SO₄)₃. 1 mole
H₂: 3 moles

The molar mass of the compounds is:

Al: 27 g/mole
H₂SO₄: 98 g/mole
Al₂(SO₄)₃: 342 g/mole
H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 2 moles ×27 g/mole= 54 grams
H₂SO₄: 3 moles ×98 g/mole= 294 grams
Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

$mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}$

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield} x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know: