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4 years ago

What is the difference between elements, mixtures, and compounds?

Chemistry

1 answer:

MAXImum [283]4 years ago

6 0

Answer:

If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.

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One type of electromagnetic radiation which has more energy than infrared

aniked [119]

Answer:

The different types of radiation are defined by the the amount of energy found in the photons. Radio waves have photons with low energies, microwave photons have a little more energy than radio waves, infrared photons have still more, then visible, ultraviolet, X-rays, and, the most energetic of all, gamma-rays.

Explanation:

8 0

3 years ago

Read 2 more answers

72.0 grams of water how many miles of sodium with react with it?

Flura [38]

Answer:

$\large \boxed{\text{8.00 mol}}$

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ: 18.02

2Na + H₂O ⟶ 2NaOH + H₂

m/g: 72.0

2. Moles of H₂O

$\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O}} = \text{3.996 mol H$_{2}$O}$

3. Moles of Na

The molar ratio is 2 mol Na/1 mol H₂O.

$\text{Moles of Na} = \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}$

7 0

3 years ago

The boiling point of an aqueous solution is 101.02 °C. What is the freezing point?

kompoz [17]

Colligative properties calculations are used for this type of problem. Calculations are as follows:

ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
ΔT(boiling point) = (Kb)m
m = 1.02 °C / 0.512 °C kg / mol
m = 1.99 mol / kg

ΔT(freezing point) = (Kf)m
ΔT(freezing point) = 1.86 °C kg / mol (1.99 mol / kg)
ΔT(freezing point) = 3.70 °C
Tf - T = 3.70 °C
T = -3.70 °C

3 0

4 years ago

If 3.5 moles of nitrogen monoxide (NO) react with 6.0 moles of oxygen gas (O2), how many moles of the product can be formed and

jasenka [17]

Answer: -

4.25 mol of O₂ left as excess.

3.5 mol of NO₂ formed.

Explanation: -

Number of moles of NO taken = 3.5

Number of moles of O₂ taken = 6.0

The balanced chemical equation for this reaction is

2 NO+ O₂ → 2 NO₂

From the equation we can see that

2 mol of NO react with 1 mol of O₂

3.5 mol of NO react with $\frac{1 mol O2}{2 mol NO}$ x 3.5 mol NO

= 1.75 mol O₂

So Oxygen O₂ is in excess and NO is the limiting reagent.

Moles of O₂ left over = 6 - 1.75 =4.25 mol of O₂

From the balanced chemical equation we see

2 mol of NO gives 2 mol of NO₂

3.5 mol of NO gives $\frac2 mol NO2}{2 mol NO}$ x 3.5 mol NO

= 3.5 mol of NO₂

5 0

3 years ago

After decaying for 48 hours, l/16 of the original mass of a radioisotope sample remains unchanged. What is the half-life of this

ivolga24 [154]

After the first half-life, 1/2 is unchanged. After the 2nd one, 1/4 remains. After the 4th one, 1/16 remains. So you have 4 half-lives in 48 hours, the half-life is thus 12 hours.

3 0

3 years ago