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3 years ago

Explain why elements and compounds are pure substances

2 answers:

8 0

N2 - Element; Pure substance O2 - Element; Pure substance N2O - Compound; Pure Substance Air - Homogeneous; Solution

neonofarm [45]3 years ago

6 0

Elements and compounds are substances because we know exactly what is in them. Elements are only one kind of atom and have a symbol. Compounds have a formula telling us exactly how many of what kind of atom make up that compound

Solutions, homogeneous mixtures, not only contain 2 compounds, but we do not know how much of each. We can make salt water with one grain of salt in a cup of water or a drop of water in a cup of salt. Mixtures do not have a formula because they can be any concentration.

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What is one property that is different between water and oil? *

Arlecino [84]

Oil is less dense than water, so the difference would be its density. Water is a good solvent, which means It can dissolve other substances.

3 0

3 years ago

Which situation would lead to increased competition

I am Lyosha [343]

I would say B because c and d would decrease competition and a would do the same, or just kill the ecosystem.

Hope this helps and don't forget to hit that heart :)

8 0

3 years ago

calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure

andriy [413]

Answer:

5.5 L

Explanation:

First we convert 10 g of propane gas (C₃H₈) to moles, using its molar mass:

10 g ÷ 44 g/mol = 0.23 mol

Then we use the PV=nRT formula, where:

P = 1 atm & T = 293 K (This are normal conditions of T and P)

n = 0.23 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

V = 5.5 L

3 0

3 years ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

$24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of 0.541284 moles of MgO:

$m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}$ .

7 0

3 years ago

If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f

I am Lyosha [343]

Answer:

0.032 L or 32 mL

Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 9.0 M

V1 = This is what we're looking for.

M2 = 0.145 M

V2 = 2 L

Solve for V1 --> V1 = M2V2/M1

V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L

3 0

3 years ago