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Cerrena [4.2K]
1 year ago
11

In the gravimetric method, is it important that you dissolve your sample in exactly 10 ml of water? why or why not?.

Chemistry
1 answer:
alekssr [168]1 year ago
8 0

When utilizing the gravimetric method, it is crucial to completely dissolve your sample in 10 mL of water. A quantitative technique called gravimetric analysis employs the selective precipitation of the component under study from an aqueous solution.

A group of techniques known as gravimetric analysis are employed in analytical chemistry to quantify an analyte based on its mass. Gravimetric analysis is a quantitative chemical analysis technique that transforms the desired ingredient into a substance (of known composition) that can be extracted from the sample and weighed. This is a crucial point to remember.

Gravimetric water content (g) is therefore defined as the mass of water per mass of dry soil. To calculate it, weigh a sample of wet soil, dry it to remove the water, and then weigh the dried soil (mdry). Dimensions of the sample Water is commonly forgotten despite having a density close to one.

To know more about gravimetry, please refer:

brainly.com/question/18992495

#SPJ4

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For each of the following compounds, identify what type of bonding holds them together.
svet-max [94.6K]

Explanation

NaCl:  Ionic crystal lattice forces

Hg:   Metallic bonding

CO₂:  London dispersion forces

CH₄:  London dispersion forces

Li₂O:  Ionic crystal lattice forces

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Ionic crystal lattice forces are strong electrostatic force of attraction between oppositely charged ions arranged into a crystal lattice of ionic compound. NaCl and Li₂O are ionic compounds

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4 0
3 years ago
An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not
Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

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2 years ago
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