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Ratling [72]
4 years ago
13

Which of the following factors affect friction?

Chemistry
1 answer:
Sunny_sXe [5.5K]4 years ago
8 0

Answer:

A

Explanation:

surface on which the object is moved or the nature of the surface .i.e. rough surface ,smooth surface ,liquids

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How is a radioactive element different from a stable element?
USPshnik [31]
A radioactive element has an unstable nucleus that emits particles in the form of alpha, beta, or gamma radiation. A stable element has a nucleus that does not emit such particles
8 0
3 years ago
Which elements are<br> considered "Noble Metals"?
Lisa [10]

Answer:

ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).

Explanation:

7 0
3 years ago
What is the current produced when a 12-Volt battery encounters a resistance of 20 Ohms?
Verdich [7]

Answer:

Explanation:

V = IR

I = V/R

I = (12V)/(20Ω)

I = 0.6 A

4 0
3 years ago
In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc
natita [175]

Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water 

x= 0.7535 g water




3 0
3 years ago
If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

                                 3H2 + N2 ------> 2NH3

from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

                                 3H2 + N2 ------> 2NH3

from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0
3 years ago
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