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LenaWriter [7]
3 years ago
5

The following ionic compounds are found in common household products. write the formulas for each compound:

Chemistry
1 answer:
victus00 [196]3 years ago
3 0

Answer:

A. Potassium phosphate: K3PO4

B. Copper (II) sulfate: CuSO4

C. Calcium chloride: CaCl2

D. Titanium dioxide: TiO2

E. Ammonium nitrate: NH4NO3

F. Sodium bisulfate: NaHSO4

Explanation:

All are ionic salts with different valent (mono valent cation or poly valent cation and anions) ions.

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Peter and Jessica are both forensic scientists. Peter visits crime scenes to pr
ratelena [41]

Answer: Peter is a Field Officer and Jessica is a Lab Officer

Explanation:

For Plato it should be, Peter is a "Field officer" because he's going into the fiend of the crime scene to find his evidence

And Jessica is a "Lab officer" because she stay's in the lab to analyze the evidence brought to her by Peter  

8 0
3 years ago
Why would you add boiling chips/stones to a solution that is to be refluxed? when should you add them?
anygoal [31]
Answer: Boiling chips provide surfaces on which bubbles can form as the liquid boils.

Hope this helps!
6 0
3 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
The chemical equation shown represents photosynthesis. Carbon dioxide plus A plus light with a right-pointing arrow towards B pl
astra-53 [7]

Answer: <u><em>Option B; It traps light energy and converts it into chemical energy.</em></u>

<h2>Explanation: This substance is chlorophyll. It is a pigment present in leaves of all plants. It absorbs light energy and provides it to carry out the process of photosynthesis. Light energy is converted into chemical energy, in form of NADPH and ATP, which can be used by plants for photosynthesis.</h2><h2></h2><h2>This pigment is present only in plants, so option A is incorrect.</h2><h2></h2><h2>This pigment only absorbs and transfers energy to other molecules, and is not associated with carbon dioxide directly, so option C and D are also incorrect.</h2>

3 0
2 years ago
1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm
sashaice [31]
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm


Ideal gas equation: pV = nRT => n = pV / RT


R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm


=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol


2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type


X (+) + O2 (g) ---> X2O          or   


2 X(2+) + O2(g) ----> X2O2 = 2XO     or


4X(3+) + 3O2(g) ---> 2X2O3


 
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)



So, lets probe those 3 cases.


3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X


Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol


That does not correspond to any of the metal with valence 1+


So, now probe the case 2.



4) Case 2:


2moles X metal / 1 mol O2(g) = x / 0.01226 mol


=> x = 2 * 0.01226 = 0.02452 mol


And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol


That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.


4) Case 3


4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635 


atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol


That does not correspond to any metal.


Conclusion: the identity of the metallic element could be titanium.
5 0
3 years ago
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