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3 years ago

Chromosomes _____. (Choose all that apply)

Chemistry

1 answer:

balu736 [363]3 years ago

4 0

Answer:

1.) Are tightly coiled bundles of DNA

2.) Carry many genes

3.) Are found inside the nucleus of eukaryote

4.) Can be used to diagnose genetic disorders

Explanation:

A chromosome is a thread-like structure made up of DNA. The chromosomes are made of DNA which are tightly coiled around protein.

Chromosomes carry hundreds and thousands of genes. The number of genes in a cell outnumbers the number of chromosomes.

In eukaryotic cells, all the chromosomes are found inside the nucleus of eukaryote.

The chromosomes and DNA can be examined so as to diagnose genetic disorders.

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Each step in the following process has a yield of 70.0%. CH4+4Cl2⟶CCl4+4HCl CCl4+2HF⟶CCl2F2+2HCl The CCl4 formed in the first st

Nadusha1986 [10]

Answer:

n(HCl)=1.96 mol

Explanation:

CH4+4Cl2⟶CCl4+4HCl

CCl4+2HF⟶CCl2F2+2HCl

With ideal yields we will end up with 4 moles of HCl.

With 70% yields on every stage

n(HCl)=0.7*0.7*4=1.96 mol

8 0

3 years ago

What is the percent composition by mass of hydrogen in nh4hco3

Simora [160]

Answer: The percent composition by mass of hydrogen in given compound is 6.33 %

Explanation:

We are given:

A chemical compound having chemical formula of $NH_4HCO_3$

It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms

To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

$\%\text{ composition of Hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of compound}}\times 100$

Mass of compound = $[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol$

Mass of hydrogen = $(5\times 1)=5g/mol$

Putting values in above equation, we get:

$\%\text{ composition of Hydrogen}=\frac{5g/mol}{79g/mol}\times 100=6.33\%$

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %

6 0

3 years ago

If the chemist mistakenly makes 250 mL of solution instead of the 200 mL, what molar concentration of sodium nitrate will the ch

Evgesh-ka [11]

Answer:

0.120M is the concentration of the solution

Explanation:

Assuming the mass of sodium nitrate dissolved was 2.552g

Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.

To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:

Moles NaNO3 -Molar mass: 84.99g/mol-

2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3

Molar concentration:

0.0300 moles NaNO3 / 0.250L =

<h3>0.120M is the concentration of the solution</h3>

7 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago

Concentration of 10.00 mL of HBr if it takes 5 mL of a 0.253 M LiOH solution to neutralize it?

Sonbull [250]

In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M

The concentration of HBr is 0.127 M.

3 0

3 years ago