The part of an experiment used to compare all the other groups

Explain the differences between bronze and brass, and steel and cast iron. Which is best used for cooking wares?

Dennis_Churaev [7]

Answer:

Steel and cast iron

Explanation:

They are all metal but assuming that you are finding the best material for your pan i suggest going for steel or cast iron

8 0

2 years ago

In a thin layer chromatography experiment, a plate of length 9.3 cm was used and a horizontal line was made at 1.45 cm above the

dezoksy [38]

Answer: The $R_f$ value is 0.664

Explanation:

Distance travelled by solvent front = (7.7-1.45)cm = 6.25 cm

Distance travelled by unknown = (5.6-1.45) cm = 4.15 cm

The retention factor or the $R_f$ value is defined as the ratio of distance traveled by the unknown to the distance traveled by the solvent front.

$R_f=\frac{\text {distance travelled by unknown}}{\text {distance travelled by solvent}}$

$R_f=\frac{4.15}{6.25}=0.664$

Thus the $R_f$ value is 0.664

6 0

2 years ago

PLZ HELP PLZ PLZ ILL MARK AS BRAINLIESTT!!!!

LuckyWell [14K]

Q.1-

Given,

mass - 10grams

volume - 24 cm³

density = mass/volume

density = 10/24

density = 0.416 g/cm³

Q.2-

Given,

mass - 700grams

volume - 1100cm³

density = mass/volume

density = 700/1100

density = 0.6363 g/cm³

5 0

2 years ago

A 2.5 L flask is filled with 0.25 atm SO3, 0.20 atm SO2, and 0.40 atm O2, and allowed to reach equilibrium. Assume at the temper

Anit [1.1K]

Explanation:

Reaction equation for the given chemical reaction is as follows.

$2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}$

Equation for reaction quotient is as follows.

Q = $\frac{P^{2}_{SO_{2}} \times P_{O_{2}}}{P^{2}_{SO_{3}}}$

= $\frac{(0.20)^{2} \times 0.40}{(0.25)^{2}}$

= 0.256

As, Q > K (= 0.12)

The effect on the partial pressure of $SO_{3}$ as equilibrium is achieved by using Q, is as follows.

This means that there are too much products.

Equilibrium will shift to the left towards reactants.

More $SO_{3}$ is formed.

Partial pressure of $SO_{3}$ increases.

4 0

2 years ago

A sample of a vapor occupies a volume of 500 mL at 65°C. If pressure remains constant, what is the volume of the gas at standard

DedPeter [7]

Answer:

403 mL

Explanation:

First, I will assume that the mole is 1, because you are not specifing this.

Now, with the innitial data, we need to get the pressure:

T = 65+273 = 338 K

V = 500 / 1000 = 0.5 L

Now if:

PV = nRT

Then:

P = nRT/V and V = nRT/P

Let's calculate the P:

P = 1 * 0.082 * 338 / 0.5 = 55.432 atm

The standard temperature is 0° C or 273 K so, the volume is:

V = 1 * 0.082 * 273 / 55.432

V = 0.40384 L or simply 403.84 mL

8 0

2 years ago