Answer:
I can use a dichotomous key. It helps me classify objects by sorting it out with "yes" and "no" questions.
or
I can use a Punnett Square. It helps me classify what genes the offspring will receive simply by figuring out the recessive and dominant genes as well as the hetzygous and homzygous.
Now give an example of which ever chart you choose by drawing it if that is required. For the Punnett Square label each of the squares Top right Hetzygous, top left dominant, bottom left recessive, bot-tom right homzygous. And for the dichotomous key put a 5-7 length branch showing the animals that have fur, can breathe under water, what cannot or doesn't have those traits. or something similar
Hopefully this helps :)
Answer:
Mass percentage → 0.074 %
[F⁻] = 741 ppm
Explanation:
Aqueous solution of flouride → [F⁻] = 0.0390 M
It means that in 1L of solution, we have 0.0390 moles of F⁻
We need the mass of solution and the mass of 0.0390 moles of F⁻
Mass of solution can be determined by density:
1g/mL = Mass of solution / 1000 mL
Note: 1L = 1000mL
Mass of solution: 1000 g
Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g
Mass percentage → (Mass of solute / Mass of solution) . 100
(0.741 g / 1000 g) . 100 = 0.074 %
Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)
0.741 g . 1000 mg/1g = 741 mg
1000 g . 1 kg/1000 g = 1kg
741 mg/1kg = 741 ppm
1 mole=6.02 ×10
8.3 moles=?
8.3×6.02×10=499.66 molecules
therefore ghe no. of molecules = 499.66
a. 0.137
b. 0.0274
c. 1.5892 g
d. 0.1781
e. 5.6992 g
<h3>Further explanation</h3>
Given
Reaction
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
2.46 g of water
Required
moles and mass
Solution
a. moles of water :
2.46 g : 18 g/mol = 0.137
b. moles of butane :
= 2/10 x mol water
= 2/10 x 0.137
= 0.0274
c. mass of butane :
= 0.0274 x 58 g/mol
= 1.5892 g
d. moles of oxygen :
= 13/2 x mol butane
= 13/2 x 0.0274
= 0.1781
e. mass of oxygen :
= 0.1781 x 32 g/mol
= 5.6992 g
