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Phantasy [73]
2 years ago
11

This problem has been solved! See the answer Determine Um (mode ), average U, and Urms for a group of ten automobiles clocked by

radar at speeds 38,44,48,50,55,55,57,58 and 60mi/h respectively
Chemistry
1 answer:
lara31 [8.8K]2 years ago
4 0

Explanation:

The mode is the most common number.

Um = 55

The mean is the sum of the numbers divided by the quantity.

Uavg = (38 + 44 + 45 + 48 + 50 + 55 + 55 + 57 + 58 + 60) / 10

Uavg = 51

The RMS (root mean square) is the square root of the sum of the squares of the numbers divided by the quantity.

Urms = √[(38² + 44² + 45² + 48² + 50² + 55² + 55² + 57² + 58² + 60²) / 10]

Urms = 51.451

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At 10.°C, 20.g of oxygen gas exerts a pressure of 2.1atm in a rigid, 7.0L cylinder. Assuming ideal behavior, if the temperature
Alja [10]

Answer:

Final pressure = 2.3225 atm

Amontons’s law states that

At constant volume and number of molecules, the pressure of a given mass of gas is directly proportional to its temperature

Explanation:

Temperature causes increased excitement of gas molecules increasing the number of collisions with the walls of the container which is sensed as increase in pressure

Amontons’s law: P/T = Constant at constant V and n

That is P1/T1 = P2/T2

Where temperature is given in Kelvin

Hence T1 of 10°C = 273.15 + 10 = 283.15K

Also temperature T2 of 40°C = 313.15 K

Hence

P2 = (P1/T1)×T2 = (2.1/283.15)×313.15 = 2.3225 atm

3 0
2 years ago
Stabiliţi numerele de oxidare ale tuturor elementelor prezente în următoarele substanţe chimice, ţinând cont de principalele reg
natita [175]

Answer:

a.

N = +2

O = -2

b.

Na = +1

O = -2

N = +5

c.

O = -2

Al = +3

P = +5

d.

O = -2

Ca = +2

C = +4

e

O = -2

Mn = +7

K = +1

f.

O = -2

K = +1

Cr = +6

g.

O = -2

Cl = +7

Cr = +1

h.

O = -2

H = +1

Cr = +5

i.

O = -2

H = +1

Cr = +3

k.

O = -2

H = +1

Cr = +1

Explanation:

A. NU

Numărul de oxidare a azotului = +2

Numărul de oxidare a oxigenului = -2

b. NaNO₃

Numărul de oxidare de Na = +1

Numărul de oxidare de O = -2

Numărul de oxidare de N = +5

c. AlPO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Al = +3

Numărul de oxidare al P = +5

d. carbonat de calciu

Numărul de oxidare de O = -2

Numărul de oxidare de Ca = +2

Numărul de oxidare de C = +4

e. KMnO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Mn = +7

Numărul de oxidare al lui K = +1

f. K₂Cr₂O₇

Numărul de oxidare de O = -2

Numărul de oxidare al lui K = +1

Numărul de oxidare al Cr = +6

g. HClO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Cl = +7

Numărul de oxidare al Cr = +1

h. HClO₃

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +5

i. HClO₂

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +3

k. HClO

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +1.

6 0
1 year ago
Ethene is a useful substance that can form polymers. It has a melting point of 169°C and a boiling point of 104°C. At which temp
blondinia [14]

Answer:

-169°C to -104°C

Explanation:

Ethene, also known as ethylene exists in solid, liquid and gaseous states. Ethene is an aliens with condensed structural formula C2H4. Athens is a colourless gas. It is flammable and is also a sweet smelling gas in its pure form. It is the monomer in the production of polyethylene which is of great importance in the plastic industry. In agriculture, it is used to induce the ripening of fruits. It can be hydrated in order to produce ethanol.

The liquid range of ethene refers to the temperatures at which ethene is found in the liquid state of matter. It is actually the difference between the melting point and the boiling points of ethene. Hence the liquid range of ethene is -169°C to -104°C

4 0
2 years ago
One reactant with a mass of 10 grams is combined with another reactant with a mass of 8 grams in a sealed container. After the r
disa [49]
18g is the most reasonable mass after the reaction
3 0
2 years ago
What volume will 50.2 grams of co2 (g) occupy at stp?
Genrish500 [490]
Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols

For every 1 mol of gas, there will be
24000 cm^3 of gas

Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
8 0
2 years ago
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