Ask question

Ask question

All categories

3 years ago

5

A dogfish shark contains 24 chromosomes in each of its muscle cells. How many chromosomes are normally found in each of its game

tes

1 answer:

Marina CMI [18]3 years ago

3 0

Well in each gamete there are 23 chromosomes

You might be interested in

Which voice part sings the lowest notes in an SATB choir?

Talja [164]

Hello

Question: which voice sings lowest in a SATB choir?

Answer: A. Bass

Hope this helps and have a great day!

4 0

3 years ago

PLEASE HELP!!!!!!!<br> WILL GIVE BRAINLIEST!!!!!<br> Describe a volcanic eruption.

Citrus2011 [14]

Answer:

A volcanic eruption occurs when molten rock, ash and steam pour through a vent in the earth's crust. Volcanoes are described as active (in eruption), dormant (not erupting at the present time), or extinct (having ceased eruption; no longer active).

Please mark as brainliest please

4 0

3 years ago

What is the difference between a 1s orbital and a 2s orbital is that

iVinArrow [24]

Answer:

The main difference is their energy level, 2s orbital is higher than 1s orbital.

4 0

3 years ago

Dose your heart muscle get tired?<br> Yes or no.

WARRIOR [948]

Answer:

<em>no</em><em> </em>

Explanation:

our heart muscles never get tired, because it has to pump blood in our body 72 times a minute, it is made of special cardiac muscles which helps it to perform it's function without getting tired ....

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

5 0

2 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago