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jarptica [38.1K]
3 years ago
11

From a cross between a short-haired guinea pig (hh) and a long-haired guinea pig (Hh), what would be the possible phenotypes of

the offspring?
A. 100% will have long hair.
OB. 75% will have long hair; 25% will have short hair.
OC. 50% will have long hair; 50% will have short hair.
O D. 25% will have long hair; 75% will have short hair.
Chemistry
1 answer:
OleMash [197]3 years ago
5 0

Answer:

C

Explanation:

There will be a two long hair and two short hair

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After performing a dilution calculation, you determine you need 25.0 milliliters of an aqueous stock solution to make 100.0 mill
IRINA_888 [86]
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.

D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.

C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.

B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.

I would add the final step of stirring, but B is the only answer that can be correct.
8 0
3 years ago
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Hydrogen's sole electron occupies the 1s orbital but can be excited to the 4p orbital. List all the orbitals this electron might
vesna_86 [32]
4p, 3d, 3p, 3s, 2p, 2s and 1s orbitals may be occupied during de-excitation.<span />
8 0
3 years ago
Which of the following molecules experience dipole-dipole forces as its strongest IMF? A) H2 B) SO2 C) NH3 D) CF4 E) BCl3
Xelga [282]

Answer: NH_3

Explanation:

a) H_2: This  is a non polar covalent compound which are held by weak vanderwaal forces of attraction.

b) SO_2: This  is a covalent compound which is polar due to the presence of lone pair of electrons and are held by dipole-dipole forces of attraction.

c) NH_3: These are joined by a special type of dipole dipole attraction called as hydrogen bond. It forms between electronegative nitrogen atom and hydrogen atom and is the strongest interaction.

d) CF_4: This  is a covalent compound and is non polar which are held by weak vanderwaal forces of attraction.

e)  BCl_3: This  is a covalent compound and is non polar which are held by weak vander waal forces of attraction.

5 0
3 years ago
Bela made scrambled eggs while camping. She placed the eggs in a pan on the grill, then turned on the flame to apply heat to the
Alina [70]

Answer: Temperature.

7 0
2 years ago
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Determine what mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the reaction CO(g) +
siniylev [52]

Answer:

5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.

Explanation:

The balanced reaction:

CO (g) + 2 H₂ (g) -> CH₃OH (l)

By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:

  • CO: 1 mole
  • H₂: 2 moles
  • CH₃OH: 1 mole

Being the molar mass of each compound:

  • CO: 28 g/mole
  • H₂: 1 g/mole
  • CH₃OH: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CO: 1 mole* 28 g/mole= 28 grams
  • H₂: 2 moles* 1 g/mole= 2 grams
  • CH₃OH: 1 mole* 32 g/mole= 32 grams

Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:

  • If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?

mass of carbon monoxide=\frac{6000 grams of methanol*28 grams of carbon monoxide}{32 grams of methanol}

mass of carbon monoxide= 5250 grams= 5.25 kg

If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?

mass of hydrogen=\frac{6000 grams of methanol*2 grams of hydrogen}{32 grams of methanol}

mass of hydrogen= 375 grams

<u><em>5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol. </em></u>

8 0
2 years ago
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