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3 years ago

What is the genetic makeup of both parents ?

Chemistry

1 answer:

Effectus [21]3 years ago

7 0

Answer:

A, T, C y G, son las "letras" del código del ADN; representan los compuestos químicos adenina (A), timina (T), citosina (C) y guanina (G), respectivamente, que constituyen las bases de nucleótidos del ADN. ...El código genético es el conjunto de reglas que define cómo se traduce una secuencia de nucleótidos en el ARNm a una secuencia de aminoácidos en una proteína

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15.

lorasvet [3.4K]

C.

Because the spinal cord functions primarily in the transmission of neural signals between the brain and the rest of the body

5 0

3 years ago

Two protons and two neutrons are released as a result of this reaction. mc017-1.jpgRn mc017-2.jpgPo + ? Which particle is releas

Dovator [93]

The nuclear reaction occurring is known as alpha-decay, and during this process, an alpha particle is released from a heavy radioactive nucleus to form a lighter more stable nucleus. The alpha particle is equivalent to a helium nucleus, which means it contains 2 protons and two neutrons (net charge of +2)
The decay equation is:
Rn → Po + α

5 0

3 years ago

Read 2 more answers

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

Bromine (Br) belongs to the halogen family. Based on the location of the halogens, how many valence electrons does bromine have?

Amanda [17]

Answer:

D. 7

Explanation:

The halogens are found in Group 7 of the Periodic Table. If Bromine is a halogen, then that means it would be found in Group 7. Also, elements in Group 7 have 7 electrons in its outer shell, Bromine would have 7 electrons in its outer shell.