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2 years ago

Describe How did the sun form?

2 answers:

8 0

The Sun formed about 4.6 billion years ago in a giant, spinning cloud of gas and dust called the solar nebula. As the nebula collapsed under its own gravity, it spun faster and flattened into a disk.

babunello [35]2 years ago

4 0

Answer:

<h2>in space billons of year ago dust particles together to create solar nebula also the sun made with helium and hydrogen the sun will run out of energy</h2>

hope it's helpful thanks

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How many mg of O2 will react with the TNT?

Tanzania [10]

TNT has the molecular formula: C7H5N3O6. And hence, when reacted in oxygen gas, you get what is known as combustion reaction. the reaction is: C7H5N3O6+O2→CO2+N2+H2O

6 0

3 years ago

A 0.02 molar solution of sodium chloride contains 0.1 mole of solute. What is the volume of the solution

Deffense [45]

Answer:

Volume of solution = 5 L

Explanation:

Given data:

Molarity of solution = 0.02 M

Moles of solute = 0.1 mol

Volume of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

by putting values,

0.02 M = 0.1 mol / volume of solution

Volume of solution = 0.1 mol / 0.02 M

Volume of solution = 5 L

3 0

2 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

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8 0

2 years ago

You have 180 g of radium that takes up 36 cm ^ 3 of space if you were to cut it in half and have only 90 g

Leno4ka [110]

If you were to cut the radium in half and have only 90 g, it will take up 18 cm³.

<h3>What is density?</h3>

The density of an object is the ratio of mass to volume of object.

Density = mass/volume

volume = mass/density

at a constant density, the volume of an object is proportional to its mass.

From the question, you have 180 g of radium that takes up 36 cm ^ 3 of space if you were to cut it in half and have only 90 g, the new mass will take the following volume.

180 g = 36 cm³

90 g = ?

= (90 x 36) / 180

= 18 cm³

Thus, if you were to cut the radium in half and have only 90 g, it will take up 18 cm³.

Learn more about radium here: brainly.com/question/23781489

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6 0

1 year ago

g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.

Lady bird [3.3K]

Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

6 0

3 years ago