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3 years ago

3. What as one of the surface process

1 answer:

6 0

I don’t know the answer question but I don’t know what about the answer question but please answer question question mark on the answer question question mark on the answer question

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An unknown solution has a pH of 7. How would you classify this solution?

xenn [34]

Solution with a ph of 7 Is neutral

7 0

3 years ago

Kindly tell me all the important things to remember about periodic table. Like, how it was made, it's history and other stuff. T

Molodets [167]

Remember that the:
1. Vertical columns are groups and families
2. Horizontal columns are periods
3. Period 1 has 2 e-, 2 has 8 e-, 3 has 8 e-, and 4 has 18 e-
3.The number on top of the groups/families shows what group it is on (ex. 1A, 2A, 3B, 4B, etc) and the number on top of that indicates the number of the same thing (what group it is)
⇒ ex. 1A has 1 val e-. 2A has 2 val e-. 3A has 3 val e- ( and rest goes up until 8A). Note: I skipped the transitions metals since they don't have accurate amount of val e-.
4. Know which is the mass number, atomic number, symbol, or name.
5. Where the metals, non- metals, and metalloids are located.
6. Elements on the same groups have similar properties.
7. The first group is Akali Metals and are very unstable and very reactive
8. Second group is Akaline Earth Metals
9. The middle, groups 3-12 are Transition Metal
10. Group 13 is the Boron Family.
11. Group 14 is the Carbon Family
12. Group 15 is the Nitrogen Family
13. Group 16 is the Oxygen Family.
14. Group 17 is the Halogen, salt families
15. Group 18 is the Noble Gases, and are very stable and not reactive.
I hope this helps. :)

7 0

3 years ago

Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be

serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

3 CuSO₄ + 2 Al ⇒ 3 Cu + Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

160 g of CuSO₄ ----------------- 1 mol

156 g ----------------- x

x = (156 x 1) / 160

x = 0.975 moles

27 g of Al -------------------------- 1 mol

29 g of Al -------------------------- x

x = (29 x 1)/27

x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical 3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental 0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.

3 moles of CuSO₄ ------------------ 3 moles of Cu

0.975 moles of CuSO₄ --------------- x

x = (0.975 x 3)/3

x = 0.975 moles of Cu were obtained.

3 0

3 years ago

Read 2 more answers

What is sp² hybridisation ?

Neko [114]

Answer: the mixing of ones s and two p atomic orbitals

Explanation: hope it helps ;)

6 0

3 years ago

How much heat (in kj) is evolved in converting 1.00 mol of steam at 130.0 ∘c to ice at -55.0 ∘c? the heat capacity of steam is 2

nevsk [136]

The heat will be evolved in following process:

1) cooling steam from $130^{0}C$ to $100^{0}C$ = Q1

2) conversion of steam to water =Q2

3) Cooling of water to $0^{0}C[/tex=Q3]4) conversion of water to ice=Q45) cooling of ice from [tex]0^{0}C$ to $-55^{0}C$ =Q5

Let us calculate each energy one by one:

1) Q1:

Mass of water vapour = moles X molar mass = 1mol X 18 = 18 grams

heat capacity of steam is $2.01\frac{J}{g^{0}C }$

$Q_{1}=massXheat capacityXchangeintemperature=18X2.01X(130-100)=-1085.4J$

2) Q2:

for this we need heat of vaporisation of water = -40.7kJ/mol

so the heat evolved in conversion of stem to water is -40.7kJ

3) Q3

the specific heat of water is $4.18\frac{J}{g^{0}C }$

heat evolved in cooling of water will be $Q_{3}=massXspecific.heatXchange.in.temperature=18X4.184X100=7531.2J=7.531kJ$

4)Q4:

The heat of conversion of ice to water (melting) is 6.02kJ/mol

5) Q5

The cooling of ice will require

$Q_{5}=massXspecificheatXchangeintemperature=18X2.09X(-55)=-2069.1J=-2.069kJ$

The total energy released is

$Q_{total}=Q1+Q2+Q3+Q4+Q5=1.085+40.7+7.531+6.02+2.069=59.474kJ$

The heat released will be 59.474 kJ

5 0

3 years ago