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4 years ago

What science project should i do?

Chemistry

2 answers:

GalinKa [24]4 years ago

6 0

You should do which candle burn the fastest colored or white

JulijaS [17]4 years ago

5 0

You could always make a seismograph.

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A sound with more vibrations per second sounds higher than a sound with fewer vibrations per second.

Fudgin [204]

Answer:

that is a false sentence

6 0

3 years ago

Read 2 more answers

What conclusion can be draw about the heat of fusion of water from the heating curve? A) The potential energy of fusion represen

zysi [14]

The answer would be c.

6 0

3 years ago

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Determine the specific heat of a material if a 15 gram sample absorbs 75 J of heat as it is temperature increases from 15°C to 4

Roman55 [17]

Answer:

$Cp=0.2\frac{J}{g\°C}$

Explanation:

Hello,

In this case, the heat, mass, specific heat and temperature change are related by:

$Q=mCp\Delta T$

Thus, if we want to compute the specific heat we simply solve for it:

$Cp=\frac{Q}{m\Delta T}=\frac{75J}{15g*(40-15)\°C} \\\\Cp=0.2\frac{J}{g\°C}$

Best regards.

3 0

3 years ago

Read 2 more answers

Anyone? Please help

snow_tiger [21]

Answer:

The limiting reacting is O2

Explanation:

Step 1: data given

Number of moles O2 = 21 moles

Number of moles C6H6O = 4.0 moles

Step 2: The balanced equation

C6H6O + 7O2 → 6CO2 + 3H2O

Step 3: Calculate the limiting reactant

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed (21 moles).

C6H6O is in excess.

For 7 moles O2 we need 1 mol C6H6O

For 21 moles O2 we'll need 21/7 = 3 moles C6H6O

There will remain 4.0 - 3.0 = 1 mol C6H6O

Step 4: calculate products

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2

For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O

The limiting reacting is O2

7 0

4 years ago

Iodine, I2, has many uses, including the production of dyes, antiseptics, photographic film, pharmaceuticals, and medicinal soap

svetoff [14.1K]

Answer:

2I ⁻ → I₂ + 2e⁻ describes the oxidation.

Cl₂ + 2e⁻ → 2Cl ⁻ describes the reduction.

Explanation:

Oxidation-reduction reaction is the simulaneous oxidation and reduction of the substances and is represented by two half-reactions.

The oxidation half-reaction is the loss of electrons, with the consequent increase in the oxidation state by the oxidized substance.

In this case, the process that shows the loss of electrons is:

2I⁻ → I₂ + 2e⁻

That reaction shows:

Two I⁻ ions lose two electrons (one each) to be oxidized to I₂.

The change in the oxidation number is from -1 to 0.

Hence this half-reaction is the oxidation reaction.

On the other hand, the reduction half-reaction is the gain of electrons, with the consequent reduction of the oxidation state by the reduced substance.

In this case, the process that shows the gain of electrons is:

Cl₂ + 2e⁻ → 2Cl⁻

That reaction shows:

Two Cl atoms gain two electrons (one each) to be reduced to Cl⁻.

The change in the oxidation number is from 0 to - 1.

Hence, this half-reaction is the reduction reaction.

Summary:

2I ⁻ → I₂ + 2e⁻ describes the oxidation.

Cl₂ + 2e⁻ → 2Cl ⁻ describes the reduction.

4 0

3 years ago