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Lynna [10]
2 years ago
5

For eyes, the male monster's genotype was EE, and the female monster's genotype was ee. What kind of eyes can the baby monsters

have?
AThey can only have two small eyes.

B
They can only have one large eye.

C
It is possible for them to have 2 small eyse or one large eye.
Chemistry
1 answer:
attashe74 [19]2 years ago
3 0

the answer that i would say is C

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How much energy does an X-ray with an 8 nm (8 x 10-9 m) wavelength have?
Vlada [557]

Answer: 2.48×10^-17 J

Explanation:

Given the following :

Wavelength = 8nm (8 x 10^-9 m)

Energy(e) of X-ray =?

Energy=[speed of light(c) × planck's constant (h)] ÷ wavelength

Speed of light = 3×10^8m/s

Planck's constant = 6.626×10^-34 Js

Wavelength = 8 x 10^-9 m

Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9

Energy = [19.878×10^(8-34)] / 8 x 10^-9

Energy = 2.48475 × 10^(-26+9)

Energy = 2.48×10^-17 J

8 0
2 years ago
What is the molarity of the solution resulting from the dissolution of 239 g glucose (C6H12O6) in 250
kifflom [539]

Answer:

Molarity =5.32 M

Explanation:

Given data:

Mass of glucose = 239 g

Volume = 250 mL (250 /1000 = 0.25 L)

Molarity = ?

Solution;

Formula:

Molarity = number of moles / volume in litter

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 239 g / 180.2 g/mol

Number of moles = 1.33 mol

Molarity:

Molarity = number of moles / volume in litter

Molarity = 1.33 mol / 0.25 L

Molarity =5.32 M

6 0
3 years ago
What is the percent composition of Fluorine in CF5?
posledela
The answer is 100%
let me know if you want an explanation
6 0
2 years ago
What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0
2 years ago
Read 2 more answers
I am running a reaction that creates 138 grams of sulfur trioxide (S03). My theoretical yield was 156 grams. What is my percenta
alex41 [277]

Answer:

88.46%

Explanation:

Percentage yield is actual/theoretical * 100

138/156 * 100 = 88.4615385

5 0
2 years ago
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