Answer:
a. .
b.
c.
Explanation:
Hello,
In this case, considering that the valve is opened, we can use the Boyle's law in order to compute the final pressure of argon by considering its initial pressure and volume and a final volume of 5.0 L:
And the final pressure of helium:
Which actually are the partial pressure of both of them, it means that the total pressure is:
Finally, the mole fraction of each gas is computed by considering the Dalton's law:
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There are two types of chemical compound one is covalent compound and other is ionic compound, covalent compound formed by sharing of electron and ionic compound formed by complete transfer of electron. Therefore, the name and formula of the compound made of magnesium and fluorine is Magnesium fluoride and MgF₂ respectively.
Chemical Compound is a combination of molecule, Molecule forms by combination of element and element forms by combination of atoms in fixed proportion.
An ionic compound is a metal and nonmetal combined compound. Ionic compound are very hard. They have high melting and boiling point because of strong ion bond.
The name and formula of the compound made of magnesium and fluorine is Magnesium fluoride and MgF₂ respectively. Magnesium fluoride is an ionic compound.
Therefore, the name and formula of the compound made of magnesium and fluorine is Magnesium fluoride and MgF₂ respectively.
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Answer:Magnesium fluorine
MgF2
Explanation:
God knows
Using Boyle's Law of gases which states that the pressure and volume of a gas have an inverse relationship when temperature is kept constant, we find that when the pressure of the gas increases from 5.0 to 7.0 atmospheres, the volume of the gas decreases to approximately 3.57 liters.
The question pertains to the application of Boyle's Law, a fundamental concept in the field of physics dealing with gases. Boyle's Law states that the pressure and volume of a gas have an inverse relationship when the temperature is held constant. This means if the pressure of a gas increases, the volume decreases, and vice versa.
In this case, you have 5.0 liters of a gas under an initial pressure of 5.0 atmospheres. The pressure is then increased to 7.0 atmospheres, and you are asked to determine the new volume of the gas. To solve this problem, we use the formula for Boyle's Law, which is P1V1 = P2V2. We know P1 (initial pressure) is 5.0 atmospheres and V1 (initial volume) is 5.0 liters. P2 (final pressure) is increased to 7.0 atmospheres and V2 (final volume) is what we are trying to find.
So, we plug the numbers into the equation and get: 5.0 atmospheres * 5.0 liters = 7.0 atmospheres * V2. Solving for V2, we find V2 to be approximately 3.57 liters. Therefore, when the pressure of the gas is increased from 5.0 atmospheres to 7.0 atmospheres, the volume decreases to around 3.57 liters, while the temperature remains constant.
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The bond angles in each of the compounds are determined by their electron-pair geometry and molecular structure.
The correct values for the indicated bond angles in each of the given compounds are:
Bond angles are determined by the electron-pair geometry and molecular structure of a compound. The given values correspond to these geometries and structures.
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The bond angles are determined by their molecular structures - SO2 is 120°, BF3 is 180°, SCl2 is 120°, CO2 is 180°, PF3 is slightly less than 109.5°, and CH4 is 109.5°.
The bond angles in various compounds are determined by the molecule's electron-pair geometry and molecular structure. For the SO2 compound, O-S-O angle corresponds to a bent molecular structure with a bond angle of 120° (electron-pair geometry: trigonal planar). The F-B-F angle in BF3 has a linear molecular structure that leads to a bond angle of 180° (electron-pair geometry: linear). In SCl2, the Cl-S-Cl angle is 120° due to its bent structure (electron-pair geometry: trigonal planar). For CO2, the O-C-O angle is 180° because of its linear structure (electron-pair geometry: linear). In PF3, the F-P-F angle is slightly less than 109.5° because of its trigonal pyramidal structure (electron-pair geometry: tetrahedral). Finally, in CH4, the H-C-H is 109.5° as it has a tetrahedral structure (electron-pair geometry: tetrahedral).
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Yes, because conservation of mass
coefficients of the reactants equal the coefficients of the products,
products and reactants are the same chemicals,
same number of each kind of atom appears in the reactants and in the products,
same number of each kind of atom appears in the reactants and in the products
Answer:
35.6 g of W, is the theoretical yield
Explanation:
This is the reaction
WO₃ + 3H₂ → 3H₂O + W
Let's determine the limiting reactant:
Mass / molar mass = moles
45 g / 231.84 g/mol = 0.194 moles
1.50 g / 2 g/mol = 0.75 moles
Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.
Let's make rules of three:
1 mol of tungsten(VI) oxide needs 3 moles of H₂
Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)
3 moles of H₂ need 1 mol of WO₃ to react
0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles
It's ok. I do not have enough WO₃.
Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.
Let's convert the moles to mass (molar mass . mol)
0.194 mol . 183.84 g/mol = 35.6 g