How do you find the amount of moles is .032 grams of water and whats the answer

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

2 Hydrogen atoms (subscript of 2) 1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

Hydrogen: 1.0079 g/mol Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

Multiply. The grams of H₂O will cancel out.

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

There are about 0.0018 moles in 0.032 grams.

In meteorology, a  cloud  is an aerosol consisting of a visible mass of minute liquid droplets, frozen crystals, or other particles suspended in the atmosphere of a planetary body. nephology is the science of  clouds, which is undertaken in the  cloud  physics branch of meteorology.cirrus clouds  usually describe clouds high in the atmosphere that can include wispy clouds, typically with ice crystals. variations include cirrus, cirrocumulus and cirrostratus as defined by the cloud's position in the atmosphere.layered stratus clouds  have both flat tops and bases, and can appear as if they take up the whole sky, stretching from horizon to horizon. other combinations and variations include stratus, stratocumulus, nimbostratus and altostratus.cumulus clouds  often pile up in multiple layers of the atmosphere, representing clouds that develop vertically. cumulus clouds often look like pillars with anvil-type tops or columns of clouds stacked vertically. variations include cumulus, cumulus-congestus, cumulonimbus and altocumulus.

The answer to this problem is. 67