In which sample are the particles arranged in a regular geometric pattern?

__________ is a set of _____ or combination of __________ alleles responsible for an _________ of a trait.

Anettt [7]

Answer:

$__________ is a set of _____ or combination of __________ alleles responsible for an _________ of a trait.$

Explanation:

7 0

3 years ago

Use the periodic table to match each of the following element symbols to its name, atomic mass, or atomic number. (3 points)

bonufazy [111]

Answer:

Tin I thunk

Explanation:

I think

8 0

3 years ago

The table describes how some substances were formed.

lisabon 2012 [21]

Explanation:

Which is a pure substance?

1. soda

2. gasoline

3. salt water

4. carbon dioxide

carbon dioxide

Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of -7.2°C. Bromine can be classified as a

1. compound.

2. impure substance.

3. mixture.

4. pure substance.

pure substance.

6 0

4 years ago

Read 2 more answers

Find the cell potential for a system whose ∆G = +55 kJ and 3 moles of electrons are exchanged.

rosijanka [135]

Answer:- cell potential = -0.19 volts

Solution:- The equation that shows the connection between $\Delta G$ and cell potential, E is written as:

$\Delta G=-nFE$

in this equation, n stands for moles of electrons, E stands for cell potential and F stands for faraday constant and it's value is $\frac{96485C}{mol}$ .

It asks to calculate the value of E, so let's rearrange the equation:

$E=-\frac{\Delta G}{nF}$

Let's plug in the values in it:

$E=-\frac{55kJ}{3mol*96485C.mol^-}$

$E=-\frac{0.00019kJ}{C}$

since, $\frac{1kJ}{C}=1000V$

Where C stands for coulombs and V stands for volts.

So, $E=-\frac{0.00019kJ}{C}(\frac{1000V}{\frac{1kJ}{C}})$

E = -0.19 V

So, the cell potential is -0.19 volts.

4 0

3 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago