Suppose you have a container filled with iron and sand. You can separate the iron from the sand if you ____________ so this is a

Chemistry

1 answer:

irinina [24]3 years ago

4 0

Answer:

Use a magnet; Mixture

Explanation:

Suppose you have a container filled with iron and sand. You can separate the iron from the sand if you use a magnet so this is a mixture.

A mixture is made up of two more components that are not chemically combined.
Mixtures may be classified as homogeneous when the components are in the same phase or heterogeneous when the components are in different phases.
A mixture of sand and iron is an example of solid-solid mixture which can be separated using a magnet.
The magnet will attract the iron in the mixture leaving the sand.

You might be interested in

PLS HELP WILL GIVE BRAINLIEST

boyakko [2]

Answer:

The IUPAC structure only shows bond pairs and lone pairs. In the flouromethane structure above, there is only one bond pair and three lone pairs of electrons. Therefore there is one electron remaining, but since it doesn't not make up a pair, it is ignored in the structure but theoretically it is present.

$.$

6 0

3 years ago

Read 2 more answers

Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by

kodGreya [7K]

Answer:

2.97 × 10¹³ g

Explanation:

First, we have to calculate the biomass the is burned. We can establish the following relations:

2.47 acre = 10,000 m²
10 kg of C occupy an area of 1 m²
50% of the biomass is burned

The biomass burned in the site of 400,000 acre is:

$400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC$

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

The molar mass of C is 12.01 g/mol
1 mole of C produces 1 mole of CO₂
The molar mass of CO₂ is 44.01 g/mol

The mass of produced is CO₂:

$8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}$