what is a non-polar molecule? Atoms are made of small particles. When atoms bond together to form molecules. They share or give electrons. If the electrons are shared equally by the atoms, then there is no resulting charge making the molecule nonpolar. Polar molecules are the complete opposite.
Answer:
The level of toxins in solutions
Answer:
Mass of water = 73.08 g
Explanation:
Given data:
Mass of hydrogen = 35 g
Mass of oxygen = 65 g
Mass of water = ?
Solution:
First of all we will write the balanced chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen = mass/ molar mass
Number of moles of hydrogen = 35 g/ 2 g/mol
Number of moles of hydrogen = 17.5 mol
Number of moles of oxygen = 65 g / 32 g/mol
Number of moles of oxygen = 2.03 moles
Now we compare the moles of water with moles hydrogen and oxygen.
H₂ : H₂O
2 : 2
17.5 : 17.5
O₂ : H₂O
1 : 2
2.03 : 2× 2.03 =4.06 mol
Number of moles of water produced by oxygen are less so oxygen is limitting reactant.
Mass of water:
Mass of water = number of moles × molar mass
Mass of water = 4.06 mol × 18 g/mol
Mass of water = 73.08 g
Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
