The distance from dalton, ga to Dallas, tx is 1308 km. What is the distance in miles

What types of star can explode in a supernova

horrorfan [7]

For a star to explode as a Type II supernova, it must be at several times more massive than the sun<span> (estimates run from eight to 15 solar masses). Like </span>the sun<span>, it will eventually run out of hydrogen and then helium fuel at its core. However, it will have enough mass and pressure to fuse carbon.
i had to right an essay on this stuff.</span>

4 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

Feeding levels within an ecosystem are

alisha [4.7K]

Answer:

C. trophic levels

Explanation:

Trophic levels are the feeding positions of all organisms in a specific ecosystem. There are five main trophic levels, or feeding levels.

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Habitats are the environment in which organisms live. They are not feeding levels.

Taxonomic levels are a way of classifying species. They consist of (in order) domain, kingdom, phylum, class, order, family, genus, and species. They are not feeding levels.

Cladograms are branching diagrams that show the relationship between multiplt organisms/species. It too is not a feeding level.

Thus, is answer is C.

hope this helps!

3 0

2 years ago

An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

melisa1 [442]

570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students

7 0

3 years ago

Read 2 more answers

Determine the amount of both hydrogen and oxygen in a 500mL sample of Water.

alina1380 [7]

There would be about 1.67 x 10^25 oxygen atoms and about 3.34 x 10^25 hydrogen atoms.

3 0

3 years ago