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Nady [450]
3 years ago
7

Examples of a pure substances

Chemistry
2 answers:
KonstantinChe [14]3 years ago
8 0

Answer:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements

Explanation:

DiKsa [7]3 years ago
3 0

Answer:

Explanation:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.

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Do all objects emit electromagnetic waves?
kherson [118]

Answer:

no, not all objects emit electromagnet waves.

Explanation: because lets say the object is carpet it has no magnetic waves in it so not all objects do.

6 0
3 years ago
Which of the following chemical equations represents an oxidation-reduction reaction?
zimovet [89]
C. Represents an oxidation-reduction reaction
7 0
3 years ago
How many milliliters of 5.0 M NaOH are needed to exactly neutralize 40. milliliters of 2.0 M HCl?
zubka84 [21]

Answer:

16mL

Explanation:

Using the following formula;

CaVa = CbVb

Where;

Where

Ca = concentration/molarity of acid (M)

Va = volume of acid (mL)

Cb = concentration/molarity of base (M)

Vb = volume of base (mL)

According to the information provided in this question;

Ca (HCl) = 2M

Cb (NaOH) = 5M

Va (HCl) = 40mL

Vb (NaOH) = ?

Using CaVa = CbVb

Vb = CaVa/Cb

Vb = 2 × 40/5

Vb = 80/5

Vb = 16mL

5 0
2 years ago
Please help me with my question ​
Allushta [10]
What is the question?
7 0
3 years ago
Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib
KengaRu [80]

Answer:

<u>For methanol:</u> Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

<u>For ethanol: </u>Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

<u>For propanol: </u>Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

<u>For methanol:</u>

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g  (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)</u>

<u></u>

<u>For ethanol:</u>

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g  (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)</u>

<u></u>

<u>For propanol:</u>

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)</u>

5 0
3 years ago
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