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Tcecarenko [31]
3 years ago
12

2Ag + S --> Ag2S would be classified as a(n) _____________ reaction.

Chemistry
1 answer:
yaroslaw [1]3 years ago
3 0

Answer:

its 2 parts creating a group which would be combustion

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What is ZnCl2 name and formula
Damm [24]

Answer:

Zinc chloride is the name of chemical compounds with the formula ZnCl2 and its hydrates

4 0
3 years ago
Both equations I and II are balanced, but equation I is the correct way to write the balanced equation.
ElenaW [278]
I don’t get the question. is there a worksheet?
5 0
3 years ago
Question 5
yuradex [85]

Answer:

it will have 1 valence electron as its in the first group in the periodic table.

It will lose electrons as its easier.

The charge will be 1+ as it lost 1 .

8 0
2 years ago
I know how to do electron configuration, but I think I’m doing the rest wrong. Answers and explanations would be much appreciate
Natasha_Volkova [10]

Your answers seem great so far, except for a tiny issue: With the ionic symbols, try to get into the habit of using "+", with metals, like sodium, and try to use the integer first. So, for example, a potassium ion would be K^+, while an oxide ion would be O^2-


Let's take aluminium as an example I'll work through:

Aluminium, with it's atomic number of 13, will have an electronic configuration of 1s2 2s2 2p6 1s2 2p1. So it would have 2, 8, 3 electrons in the first three energy levels, respectively.

Usually, if an elemental atom has a valence electron (highest energy level electron) count less than 4, it almost always will lose electrons. Since aluminium has 3, it will also lose the electrons.

It loses the 3 valence electrons, and so will end up with 10 electrons.

Since the atomic number also tells how many protons it has, we know that an aluminium atom has 13 protons, which doesn't change.

Since the size of the charges of a proton and an electron are the same, with protons being positive and the electrons being negative, an aluminium ion would have a charge of +3, and the Ionic symbol would be Al^3+



Hope I helped! xx


4 0
3 years ago
The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a
mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0
3 years ago
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