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2 years ago

How might emission spectra be used in studying stars?

Chemistry

2 answers:

nordsb [41]2 years ago

5 0

Answer:

Emission spectra may be used in studying stars in order to determine what atoms makeup the individual star produces, due to the fact that each atom’s emission spectra biunique, one can observe the spectra emitted by the star and identify the atoms that are released by the light the star produces.

malfutka [58]2 years ago

4 0

Answer:

^yes

Explanation:

what they said was right

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Predict the sign for AS for each of the following systems: Water freezing Water evaporating Crystalline urea dissolving Assembly

djyliett [7]

Answer: 1. Water freezing : $\Delta S$ is -ve.

2. Water evaporating : $\Delta S$ is +ve.

3. Crystalline urea dissolving : $\Delta S$ is +ve.

4. Assembly of the plasma membrane from individual lipids: $\Delta S$ is -ve.

5. Assembly of a protein from individual amino acids: $\Delta S$ is -ve.

Explanation:

Entropy is defined as the measurement of degree of randomness in a system.

It is represented by symbol S and we can only measure a change in entropy which is given by $\Delta S$ .

If there is decrease in randomness , the sign for $\Delta S$ is -ve and If there is increase in randomness , the sign for $\Delta S$ is +ve.

1. Water freezing: Entropy decreases as we move from liquid state to to solid state and thus $\Delta S$ is -ve.

2. Water evaporating : Entropy increases as we move from liquid state to gaseous state and thus $\Delta S$ is +ve.

3. Crystalline urea dissolving : The molecules convert from solid and ordered state to aqueous phase and random state. Thus the entropy increases and thus $\Delta S$ is +ve.

4. Assembly of the plasma membrane from individual lipids: random lipids are associating to form a single large polymer and thus entropy decreases and thus $\Delta S$ is -ve.

5. Assembly of a protein from individual amino acids: random amino acids are associating to form a single large polymer and thus entropy decreases and thus $\Delta S$ is -ve.

5 0

2 years ago

HELP!

MAVERICK [17]

Answer:

Explanation:

energy cannot be created or destroyed like solar energy for instance. it is already energy because it comes from the sun and for other reason.

Hope this helps!

4 0

2 years ago

Use the equation to determine what mass of FeS must react to form 326 g of FeCl2.

iogann1982 [59]

Answer:

We need 226 grams of FeS

Explanation:

Step 1: Data given

Mass of FeCl2 = 326 grams

Molar mass FeCl2 = 126.75 g/mol

Step 2: The balanced equation

FeS + 2 HCl → H2S + FeCl2

Step 3: Calculate moles FeCl2

Moles FeCl2 = 326 grams / 126.75 grams

Moles FeCl2 = 2.57 moles

Step 4: Calculate moles FeS needed

For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl

For 2.57 moles FeCl2 we need 2.57 moles FeS

Step 5: Calculate mass FeS

Mass FeS = 2.57 moles * 87.92 g/mol

Mass FeS = 226 grams FeS

We need 226 grams of FeS

4 0

3 years ago

What is the abbreviated electron configuration of cobalt?

drek231 [11]

It's Co just look at the periodic table

7 0

3 years ago

Read 2 more answers

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

<u>Answer:</u> The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago