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Marianna [84]
2 years ago
8

Could the cuttlefish tell the difference between the blue and yellow in the lab?​

Chemistry
1 answer:
jeka942 years ago
5 0

Answer:

<em>yes</em>

Explanation:

the cuttle fish tell the difference between blue and yellow

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Which of the following are examples of matter?a. heat b. sunlight c. water d. grass e. air
OverLord2011 [107]
Sunlight? air? i’m not sure they could all be matter..
7 0
3 years ago
Magnesium hydroxide (Mg(OH)2): g/mol
Eva8 [605]

To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.

Mg = 24.305 x 1 = 24.305 amu

O = 15.9994 x 2 =31.9988 amu

H = 1.0079 x 2 = 2.0158 amu

 

Then, add all the components to get the atomic mass of the molecule.

24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu


The atomic mass is just equivalent to its molar mass.

So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.

8 0
3 years ago
Read 2 more answers
Yeast and other organisms can convert glucose (C6H12O6) to ethanol (CH3CH2OH) by a process called alchoholic fermentation. The n
NISA [10]

Answer:

8.37 grams

Explanation:

The balanced chemical equation is:

C₆H₁₂O₆     ⇒   2 C₂H₅OH (l) + 2 CO₂ (g)

Now we are asked to calculate the mass  of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.

From the ideal gas law we can determine the number of moles that the 2.25 L represent.

From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.

PV = nRT    ⇒ n = PV/RT

n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂

Moles glucose required:

0.093 mol CO₂  x  ( 1 mol C₆H₁₂O₆   / 2 mol CO₂ ) =  0.046 mol C₆H₁₂O₆

The molar mass of glucose is 180.16 g/mol, then the mass required is

0.046 mol x 180.16 g/mol = 8.37 g

5 0
3 years ago
Compute 6.28×1013+7.30×1011.
Simora [160]
6.28×1013+7.30×1011 this =13741.94
6 0
4 years ago
A compound is 57.1% oxygen and the remainder carbon. using oxygen as the standard (16 amu) and assuming a 1:1 atom ratio, calcul
Delicious77 [7]
Let's start with the amount given in percent. Let our basis be 100 grams of compound. So, that means that in this amount, 57.1 g is oxygen and 100-57.1=42.9 g is carbon. Since there is 1:1 atom ratio, it also means that moles oxygen = moles carbon.

Moles = Mass/Relative Mass
Let x be the relative mass of oxygen

57.1/16 = 42.9/x
Solving for x,
<em>x = 12.02 amu</em>
4 0
3 years ago
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