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2 years ago

We know the sun is primarily made from hydrogen and helium on the basis of its:

Chemistry

1 answer:

Aleksandr [31]2 years ago

4 0

We know the sun is primarily made from hydrogen and helium on the basis of its spectrum.

<h3>What is Sun's spectrum?</h3>

Sun's spectrum refers to the light of the Sun whose wavelength spectra can be used to identify atomic composition.

Thus spectrum (Sun's spectrum) initiates with the red light (700 nm) and ends with the blue (380 nm).

In conclusion, we know the sun is primarily made from hydrogen and helium on the basis of its spectrum.

Learn more about the Sun's spectrum here:

brainly.com/question/2021598

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How does cooler air move in relation to warmer air?

ladessa [460]

Answer:
Cool air moves down while hot air rises

7 0

3 years ago

Which of the following methods correctly describes the preparation of 1.00 L of an aqueous solution of 0.500 M NaOH?

djyliett [7]

Answer:

Place 20.0 g NaOH(s) in a flask and dilute to 1.00 L with water.

Explanation:

7 0

2 years ago

Marina CMI [18]

Answer:

Molar mass = 30.026 g/mol

Explanation:

Let Molar mass be mm

mm of CH2O = 12.01 + 2(1.008) + 16.00

= 12.01 + 2.016 + 16.00 = 30.026 g/mol

6 0

2 years ago

A 20.7167 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation mgco3(s) → mgo(s

marishachu [46]

The mass of impure $MgCO_{3}$ is 20.7167 g. The decomposition reaction is as follows:

$MgCO_{3}(s)\rightarrow MgO(s)+CO_{2}(g)$

Here, 1 mole of $MgCO_{3}$ gives 1 mole of MgO.

Molar mass of $MgCO_{3}$ and MgO is 84.31 g/mol and 40.3044 g/mol respectively.

Converting number of moles in terms of mass,

$n=\frac{m}{M}$

Here, M is molar mass.

Since, $n_{MgCO_{3}}=n_{MgO}$

Thus, $\frac{m_{MgCO_{3}}}{M_{MgCO_{3}}}=\frac{m_{MgO}}{M_{MgO}}$

On putting the values,

$\frac{m_{MgCO_{3}}}{84.31}=\frac{m_{MgO}}{40.3044}$

Rearranging,

$m_{MgO}=\frac{m_{MgCO_{3}}}{84.31}\times 40.3044$

Or,

$m_{MgCO_{3}}=\frac{m_{MgO}}{0.4780}$ ...... (1)

Let the mass of impurity be $M_{I}$ and mass of impure $MgCO_{3}$ is 20.7167 g thus,

$M_{MgCO_{3}}=(20.7167-M_{I})g$ ...... (2)

Also, mass of impure MgO is 16.8817 g thus,

$M_{MgO}=(16.8817-M_{I})g$ ...... (3)

On comparing equations (2) and (3),

$M_{MgO}=M_{MgCO_{3}}-3.835$

Putting the value of $M_{MgO}$ in equation (1),

$m_{MgCO_{3}}=\frac{M_{MgCO_{3}}-3.835}{0.4780}$

Or,

$0.4780 M_{MgCO_{3}}=M_{MgCO_{3}}-3.835$

Or,

$M_{MgCO_{3}}-0.4780M_{MgCO_{3}}=3.835$

Or,

$0.522 M_{MgCO_{3}}=3.835$

Or,

$M_{MgCO_{3}}=\frac{3.835}{0.522}=7.35 g$

Thus, magnesium carbonate present in the original sample is 7.35 g.

5 0

3 years ago

Decane (C10H22) used to produce poly(ethene). what conditions are needed?

Strike441 [17]

Cracking of hydrocarbons involves thermal decomposition.
This means that large hydrocarbon molecules break into
smaller molecules when they are heated. The hydrocarbons
are boiled and the hydrocarbon gases are either
mixed with steam and heated to a very high temperature or
passed over a hot powdered aluminium oxide catalyst.
The catalyst works by providing the hydrocarbon gases
with a convenient surface for the cracking to take place.

For example, decane (an alkane with 10 carbons)
can be cracked to produce octane and ethene.

decane octane + ethene.
C10H22(g) C8H18(g) + C2H4(g)

Octane is used as petrol.
Ethene is used in the manufacture of polymers.

Cracking an alkane produces a smaller alkane plus an alkene.
If you add up the number of hydrogen atoms in the
above reaction, you will see that there are 22 on each side.
An alkene is produced because the original alkane does
not have enough hydrogen atoms to produce two more alkanes.

hope this helps

6 0

3 years ago