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Oksana_A [137]
3 years ago
5

What is a controlled experiment?

Chemistry
2 answers:
Viktor [21]3 years ago
6 0
A- one variable is changed at time maybe
larisa [96]3 years ago
4 0
C is becsuse it just souds right
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1- What is temperature?
Katen [24]

Answer:

the sum of average kinetic molecules of the body is called temperature..

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3 years ago
What are this element’s chemical and physical properties? Explain how the chemical properties affect the physical properties of
Setler79 [48]

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What element is this. Would you mind inserting a picture or naming the element in the comments section

3 0
3 years ago
State whether the following statements are true or false (with justification). (a) 1 mol of N2 has more molecules than 1 mol of
MAVERICK [17]

Answer:

A. False.

Every substance contains the same number of molecules i.e 6.02x10^23 molecules

B. False.

Mass conc. = number mole x molar Mass

Mass conc. of 1mole of N2 = 1 x 28 = 28g

Mass conc. of 1mol of Ar = 1 x 40 = 40g

The mass of 1mole of Ar is greater than the mass of 1mole of N2

C. False.

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of Ar = 40g/mol

The molar mass of Ar is greater than that of N2.

Explanation:

7 0
3 years ago
A compound is 75.46% Carbon, 4.43% hydrogen and 20.10% Oxygen by mass. it has a molecular weight of 318.31g/mol. what is the mol
DiKsa [7]

Answer:

No.of moles of C is , n = mass/molar mass = 75.46 g / 12 (g/mol) = 6.3 moles No.of moles of H is , n' = mass/molar mass = 4.43 g / 1.0(g/mol) = 4.43 moles No.of moles of O is , n'' = mass/molar mass = 20.10 g / 16(g/mol) =1.25 moles Ratio to the no.of moles of C,H& O is 6.3 : 4.43 : 1.25 In the simple integer ratio is ( 6.3/1.25) : ( 4.43/1.25) : (1.25/1.25) 5.04 :3.5 : 1

Explanation:

5 0
3 years ago
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
4 years ago
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