Answer:
The mixture contains 8.23 g of Ar
Explanation:
Let's solve this with the Ideal Gases Law
Total pressure of a mixture = (Total moles . R . T) / V
We convert T° from °C to K → 85°C + 273 = 358K
3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L
(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles
0.756= Total moles from the mixture
Moles of Ar + Moles of H₂ = 0.756 moles
Moles of Ar + 1.10 g / 2g/mol = 0.756 moles
Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206
We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g
Cells are organized into tissues. Tissues carry out specific functions. Groups of tissues can then form organs that also have specific functions to carry out. Lastly groups of organs form organ systems that regulate many functions in a specific part of the body.
The compound solubility which will not be affected by a low pH in solution is AgBr.
<h3>What is pH?</h3>
pH is a measure of the acidity or basicity of any solution and according to the pH scale 0 to 6.9 shows the acidity, 7 is neutral and 7.1 to 14 shows the basicity of any solution.
- AgBr is sparingly soluble in water and not soluble in acids, so if we low the pH of the solution towards the acidity its solubility not affected.
- NiCO₃ is a basic salt and and shows solubility in the acidic medium so change in pH will affect its solubility.
- Co(OH)₂ it is also a basic compound and shows its solubility in the acidic medium and get affected when change in pH takes place.
- PbF₂ is a strong base and also shows solubility in the acidic medium easily, so get affected when change in pH takes place.
- In CuS, sulphide is basic ion and whole compound shows solubility in the acidic medium and get affected when low pH of solution takes place.
AgBr is not affected by a low pH in solution.
To know more about solubility, visit the below link:
brainly.com/question/23946616
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Be and F,
E.N of Fluorine = 3.98
E.N of Beryllium = 1.57
________
E.N Difference 2.41 (Ionic Bond)
For H and Cl,
E.N of Chorine = 3.16
E.N of Hydrogen = 2.20
________
E.N Difference 0.96 (Polar Covalent Bond)
For Na and O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For F and F,
E.N of Fluorine = 3.98
E.N of Fluorine = 3.98
________
E.N Difference 0.00 (Non-Polar Covalent Bond)
Result:
A polar covalent bond is formed between Hydrogen and Chlorine atoms.
Answer:
Explanation:
Hello!
In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:
Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:
Then, we obtain:
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