We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.
150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo
1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo
Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
Answer:
Option 1. 0.55 mol of C₄H₁₀
Explanation:
The reaction is:
2C₄H₁₀ + 13O₂ → 8 CO₂ + 10H₂O
This is a combustion reaction where carbon dioxide and water are produced.
We convert mass of produced water to moles → 50 g / 18 g/mol = 2.78 moles of water.
The stoichometry states that:
10 moles of water are made by 2 moles of C₄H₁₀
Therefore 2.78 moles of water will be made by (2.78 . 2) / 10 = 0.55 mol of C₄H₁₀
Enthalpy is the change in energy. A cold pack will react endothermically and enthalpy will be positive and a heat pack will have negative enthalpy. You cannot determine which has more change in enthalpy unless you measure the temperature change. Eg. a heat pack went from 25c to 35c will have greater change in enthalpy then a cold pack going from 25 to 20c because the net temperature difference is greater
We first assume that the gas is ideal which is a safe assumption to approximate the answer to the problem. Then we need to know the ideal gas equation and that is:
PV=nRT
where
P- pressure
V- volume
n-number of moles-
R- ideal gas constant
T-temperature.
Since we know that P, T and V are constant, rearranging the equation would lead to:
P/TR = n/V or the ratio of the moles of gas and volume is constant.
(3moles)/2L = (3+x)/4L
where
x is the additional moles.
Solving for x = 3 moles.
The side of each water molecule with the oxygen atom uncovered will be marginally negative.
The side of each water molecule with the hydrogen atoms uncovered will be marginally positive.
So the two Cl{-} particles will be pulled in to the biggest number of positive charges, which happen in the boxes on the upper right and lower left.
The two Na{+} particles will be pulled in to the biggest number of negative charges, which happen in the boxes on the upper left and lower right.
