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BlackZzzverrR [31]
2 years ago
11

We know that organisms pass traits onto their offspring through reproduction. So how is it

Chemistry
2 answers:
Tpy6a [65]2 years ago
6 0

Answer:

There are recessive traits and dominant traits

Explanation:

recessive traits are always hidden

dominant traits are the ones that are present in an organism

some recessive traits can be passed on to offsprings to become dominant

and some dominant traits can be passed on to become recessive traits

that's life

MArishka [77]2 years ago
5 0

Answer:

because of genetic variation.

Explanation:

because of meiosis and a process called crossing over, there is genetic variation and diversity in the world, with humans and animals.

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What's a good model for isostasy is
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Provide an appropriate alkyne starting material A and intermediate product B. Omit byproducts. The number of carbon atoms in the
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Answer:

C4H6 + 2HBr = C4H8Br2

Explanation:

Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of CnH2n−2. They are unsaturated hydrocarbons.

Furthermore, Halogenation reaction of any alkyne such as but-2-yne with any of the hydro halogenated compounds such as HBr, HCl, HF or HI will give an alkane that will have the same number of carbon atoms in the starting material and in the final product.

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3 years ago
Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is a
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Answer:

Approximately 1.876 \times 10^{-3}\; \rm mol.

Explanation:

Convert both volumes to standard units (that is: liters.)

  • 10.00 \; \rm mL = 10.00 \times 10^{-3}\; \rm L = 1.000 \times 10^{-2}\; \rm L.
  • 27.08 \; \rm mL = 27.08 \times 10^{-3}\; \rm L = 2.708 \times 10^{-2}\; \rm L.

Number of moles of \rm HCl initially present (in the 10.00\; \rm mL solution at 1.00\; \rm M.)

n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol.

Number of moles of \rm NaOH from the titration:

n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol.

\rm NaOH neutralizes \rm HCl at a 1:1 ratio:

\rm HCl + NaOH \to NaCl + H_2O.

Hence, n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol.

\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}.

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