Answer:
28.01g
Explanation:
Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.
The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .
The molecular weight of the compound CO is therefore
equal to the sum of the weight of both elements.
That’s = 12.01g + 16.00g
= 28.01g
Therefore, the molecular weight of CO is 28.01g
Answer:
None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.
Explanation:
Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.
Before balanced Left side.
Cl-2
O-8
H-2
Before balanced right side.
H-1
Cl-1
O-3
That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.
(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
Answer:
The vapor pressure of the solution is 23.636 torr
Explanation:
Where;
is the vapor pressure of the solution
is the mole fraction of the solvent
is the vapor pressure of the pure solvent
Thus,
15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl
0.67 kg of water = [(0.67*1000)/(18)]moles = 37.222 moles of H₂O
Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)
Mole fraction of solvent (water) = (37.222)/(37.222+0.261)
Mole fraction of solvent (water) = 0.993
<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm
Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm
the vapor pressure of the solution = 0.0311 atm = 23.636 torr
Answer:
3.1 x 10⁻²¹ Nm
Explanation:
When placed in an external electric filed, an electric dipole experiences a torque. and this torque is represented mathematically with the equation:
torque (τ) = dipole moment vector (P) x electric field vector (E)
τ = P. E . sin θ
where θ is the angle between the water molecule and the electric field, which in this case is 90° (because this is where the torque is maximum)
τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . sin90
τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . 1
solve for τ
τ = 3.1 x 10⁻²¹ Nm
the maximum possible torque on the water molecule is therefore 3.1 x 10⁻²¹ Nm
