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Irina-Kira [14]
3 years ago
5

You drop a rock weighing 23.2 g into a graduated cylinder that contains 55 mL. The level

Chemistry
1 answer:
snow_lady [41]3 years ago
8 0

Answer:

<h2>3.31 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 23.2 g

volume = final volume of water - initial volume of water

volume = 62 - 55 = 7 mL

We have

density =  \frac{23.2}{7}  \\  = 3.314285

We have the final answer as

<h3>3.31 g/mL</h3>

Hope this helps you

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Read 2 more answers
1 C3H8 + 5 O2 --&gt; 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many
UNO [17]

Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen.  We also know that number of mole is proportional to pressure and volume.  Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law).  So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2.  Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol.  Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent.  Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water.  Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

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3 years ago
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4 years ago
9. Calculate the specific heat capacity of titanium if a 43.56 g sample absorbs 0.476 kJ as its temperature changes from 20.5 oC
Shalnov [3]

Answer:

c = 0.528 J/g.°C

Explanation:

Given data:

Mass of titanium = 43.56 g

Heat absorbed = 0.476 KJ  = 476 j

Initial temperature = 20.5°C

Final temperature = 41.2°C

Specific heat capacity = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 41.2°C - 20.5°C

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476 J = 43.56 g × c × 20.7 °C

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8 0
3 years ago
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