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s344n2d4d5 [400]
2 years ago
3

Jane and Mark make an H-R diagram and conclude that the Sun is the brightest and hottest star within 20 light years of Earth. Do

you agree with their claim? Why or why not?
Chemistry
2 answers:
Vikki [24]2 years ago
7 0

Answer:

No

Explanation:

Sirius A is the brightest, hot star nearest to Earth next to the Sun. It is about 8.6 light years away. It has 98% mass of the Sun confined in size smaller than our Earth. It has a surface temperature of 9940 K which is much greater than the surface temperature of the Sun (5800 K).

Thus, Jane and Mark's claim is incorrect that the Sun is the brightest and the hottest star within 20 light years of the Earth. The Sun is the brightest because of its proximity but not the hottest.

ioda2 years ago
6 0
I believe they are correct because although the sun is not the brightest star in the universe, I believe it is the brightest star with 20 light years of us.
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If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T
zhuklara [117]

Answer: The actual yield of B_2O_3 is 60.0 g

Explanation:-

The balanced chemical reaction :

B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)

Mass of B_2H_6 =Density\times Volume=1.131g/ml\times 36.9ml=41.7g

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles

According to stoichiometry:

1 mole of B_2H_6 gives = 1 mole of B_2O_3

1.51 moles of B_2H_6 gives =\frac{1}{1}\times 1.51=1.51 moles of B_2O_3

Theoretical yield of B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g

Percent yield of B_2O_3= 75.7\%

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

75.7\%=\frac{\text{Actual yield}}{79.3}\times 100

{\text{Actual yield}}=60.0g

Thus the actual yield of B_2O_3 is 60.0 g

7 0
2 years ago
Suppose a soap manufacturer starts with a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic
DIA [1.3K]

Answer:

Sodium arachidate; Sodium palmitate and Sodium palmitate

Explanation:

Triglycerides are esters of fatty acids with glycerol. In triglycerides, three fatty acid molecules are linked by ester bonds to each of the three carbon atoms in a glycerol molecule. The fatty acids may be same or different fatty acid molecules. Hydrolysis of triglycerides yields the three fatty acid molecules and glycerol.

Saponification is the process by which a base is used to catalyst the hydrolysis of the ester bonds in glycerides. The products of this base-catalyzed hydrolysis of triglycerides are the metallic salts of the three fatty acids and glycerol. The salts of the fatty acids are known as soaps.

For a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic acid attached to the three backbone carbons glycerol, the saponification of the triglyceride with NaOH will yield the sodium salts or soaps of the three fatty acids as well as glycerol.

Arachidic acid will react with NaOH to yield sodium arachidate.

The two palmitic acid molecules will each react with NaOH to yield sodium palmitate.

8 0
1 year ago
Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [
serious [3.7K]

Answer: The rate of disappearance of NO_2 is 3.5\times 10^{-5}M/s

Explanation:

The given chemical reaction is:

2NO_2\rightarrow 2NO+O_2

The rate of the reaction for disappearance of NO_2 is given as:

\text{Rate of disappearance of }NO_2=-\frac{\Delta [NO_2]}{\Delta t}

Or,

\text{Rate of disappearance of }NO_2=-\frac{C_2-C_1}{t_2-t_1}

where,

C_2 = final concentration of NO_2 = 0.00650 M

C_1 = initial concentration of NO_2 = 0.0100 M

t_2 = final time = 100 minutes

t_1 = initial time = 0 minutes

Putting values in above equation, we get:

\text{Rate of disappearance of }NO_2=-\frac{0.00650-0.0100}{100-0}\\\\\text{Rate of disappearance of }NO_2=3.5\times 10^{-5}M/s

Hence, the rate of disappearance of NO_2 is 3.5\times 10^{-5}M/s

6 0
2 years ago
The name carbohydrate comes from the fact that many simple sugars have chemical formulae that look like water has simply been ad
GREYUIT [131]

Answer:

The maximum mass of water produced is  m__{H_2O}} =116 \ g

Explanation:

From the question we are told that

    The mass of  sucrose is m_s  = 200 \ g

    The chemical formula for  sucrose is  C_{12} H_{22} O_{11}

The chemical equation for the dissociation of sucrose  is

          C_{12} H_{22}O_{4} \to 12C + 11H_2O

The number of moles of  sucrose can be evaluated as

        n =  \frac{m}{Z}

Where Z is the molar mass of sucrose which has a constant value of

      Z  = 342 \ g/mol

So  

    n = \frac{200}{342}

   n =0.585

From the chemical equation one mole of sucrose produces 11 moles of water so  0.585 moles of sucrose will produce x moles of  water

Therefore

          x =  \frac{0.585 * 11}{1}

         x = 6.433 \ moles

Now the mass of water produced is mathematically represented as

         m__{H_2O}} =  x * Z__{H_2O}

 Where  Z__{H_2O} is the molar mass of water with a constant values of  Z__{H_2O}} = 18 \ g/mol

  So  

        m__{H_2O}} =  6.43* 18

       m__{H_2O}} =116 \ g

 

     

     

7 0
2 years ago
Which diastereomer of oct−4−ene yields a single meso compound, (4r, 5s)−4,5−dibromooctane on reaction with br2?
stepladder [879]
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4 0
2 years ago
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