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3 years ago

Cell and chromosome characteristics?

Chemistry

2 answers:

rusak2 [61]3 years ago

5 0

Answer:

hope the inserted image will help :)

Explanation:

sattari [20]3 years ago

5 0

Answer: Both are centered inside the nucleus.

Explanation: The characteristics between cells and chromosomes are that they both are centered inside the nucleus (which is a cell). This characteristic applies to both plant and animal cells.

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Question 3 of 25

Lisa [10]

Answer:

I thinks it Br hope this helps

4 0

2 years ago

Calculate the amount of carbon dioxid gas in 1.505x10^23 molecules of the gas.

Daniel [21]

Explanation:

We need to find the amount of carbon dioxide gas in $1.505\times 10^{23}$ molecules of the gas.

We know that, 1 mole weighs 44 gram of carbon dioxide which contains $6.022\times 10^{23}$ number of molecules. It means that, $6.022\times 10^{23}$ number of molecules present in 44 grams of carbon dioxide molecule. So, $1.505\times 10^{23}$ number of molecules present in :

$=\dfrac{1.0505\times 10^{23}}{6.022\times 10^{23}}\times 44\\\\=7.675\ \text{grams}$

Hence, 7.675 grams of carbon dioxide is present in $1.505\times 10^{23}$ molecules of the gas.

4 0

3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

Each square of the periodic table shows information about one element. The square usually shows the element's

omeli [17]

D, all of the above.

7 0

3 years ago

Which of the following processes does not facilitate containment?

Marta_Voda [28]

Enrichment of uranium-235 is done. Hence, option A is correct.

<h3>What is the meaning of containment?</h3>

The act, process, or means of keeping something within limits the containment of health costs.

Enrichment is a process of increasing the proportion of fissile isotope found in uranium ore (represented by the symbol 'U') to make it usable as nuclear fuel or the compressed, explosive core of nuclear weapons.

Hence, option A is correct.

Learn more about containment here:

brainly.com/question/3299455

#SPJ1

7 0

2 years ago

Cell and chromosome characteristics? ​

Cell and chromosome characteristics?