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3 years ago

Chromosomes _____. (Choose all that apply)

Chemistry

2 answers:

Kruka [31]3 years ago

7 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:

devlian [24]3 years ago

5 0

1 , 2, 3, & 4
Are the answers

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Which shows the formula for an Organic acid

tresset_1 [31]

Answer:
The formula of Organic acid is as follow,

R-COOH

Explanation:
The class of organic acids is called Carboxylic Acids. In above general structure, R is alkyl group and can vary. While -COOH is the functional group.
Carboxylic Acids has the tendency to loose protons and their pKa value depends upon the alkyl group. For example the pKa value of Acetic acid (R = -CH₃) is 4.7. The driving force for this acidity is the stability of carboxylate (conjugate base) due resonance. i.e

RCOOH ⇄ RCOO⁻ + H⁺
Where;
RCOO⁻ = Carboxylate Ion (Conjugate base)

3 0

2 years ago

How many grams of Co are needed to react with an excess of fe203 to produce 156.2 g FE? show your work.

Gekata [30.6K]

The reaction is given as

Fe2O3 (s)+ 3CO(g)--->3CO2(g)+ 2Fe(s)

No.of moles=mass in gram/molar mass

As for Fe mole =156.2g/55.847=2.7969~2.797

The ratio b/w CO and Fe is 3:2

Moles of CO needed= 2.797x3/2=4.1955

Mass of CO needed= 4.195mol x 28.01g/mol= 117.515g

8 0

3 years ago

What must be true of a substance with a lot

LUCKY_DIMON [66]

Answer:

A is the answer.

Explanation:

8 0

3 years ago

Read 2 more answers

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

2 years ago

I don't quite understand it

irina1246 [14]

So, you need to have same ammount of atoms on the left and on the right side of the equation. You need to count the ammount of attoms of every substance on the left, and make sure that on the right side the ammount is same. For example in the 1st one it’s 6Sn+2P4=2Sn3P4, so that you have 6atoms of Sn on the left and 6 atoms of Sn on the right, same with the P

6 0

2 years ago