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1 year ago

A certain liquid has a density of 2.67 g/ cm3. what is the mass of 30.5 ml of this liquid? (

1 answer:

4 0

Hey there!

density = 2.67 g/cm³

volume = 30.5 mL

Therefore:

Mass = density * volume

Mass = 2.67 * 30.5

Mass = 81.435 g

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How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released durin

faust18 [17]

Answer:

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

Q = n*ΔH (2)

Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = -441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is -3,535 045.kJ/mol or -3.535x10⁶ J/mol. Solving for n we have:

n = -441,045.3 / -3.535x10⁶

n = 0.1248 moles

Finally, we can calculate the grams needed with expression (3). The molar mass of pentane is 72 g/mol

m = 0.1248 * 72

m = 8.9856 g

This is the mass needed to heat 1.39 kg of water

6 0

2 years ago

Alkenes: draw the product of 1-chloro-2-ethylcyclohexene with hydrogen gas and a platinum catalyst

atroni [7]

Reacting 1-chloro-2-ethylcyclohexene with hydrogen gas using a platinum catalyst would give a product of 1-chloro-2-ethylcyclohexane.

Hydrogen gas is a reducing agent, which in this reaction, simply mean that the alkene double bond in the cyclohexene will disappear because one of the two bonds forming the double bond (in the alkene) will be connected to a hydrogen atom. The platinum catalyst is necessary to allow the reaction to proceed at a much lower (activation) energy than would have been required.

7 0

1 year ago

The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to

adelina 88 [10]

Answer:

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

The combustion stoichiometry is as follows:

C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:

MW 44 5x32 4x18 3x44

So each gram of propane is 1/44 = 0.02272 mol propane

and will need 5 x 0.02272 = 0.1136 mol oxygen

At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.

At the low pressure in the burner we can use the Ideal Gas Law

PV=nRT, or V = nRT/P

P = 1.1 x 101325 Pa = 111457 Pa

T = 195°C + 273 = 468 K

R = 8.314

and we calculated n = number of moles air = 0.54 mol

So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.

6 0

2 years ago

If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?

Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

Answer:

<span>0.2 moles O</span>

8 0

2 years ago

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Your friend looks at a piece of ice and says “Solids, like ice, have a fixed shape because the particles are not moving.” Is you

Alex_Xolod [135]

Answer:

yes

Explanation:

6 0

1 year ago

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