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adell [148]
1 year ago
5

A graduated cylinder was filled to 20.3 mL with water. A solid object weighing 73.05 g was immersed in the water, raising the me

niscus of the water to 48.1 mL. Calculate the density of the solid object in g/mL.
Chemistry
1 answer:
victus00 [196]1 year ago
3 0

The density of the solid object will be 2.63 g/mL

<h3>What is density?</h3>

Density of objects = mass/volume.

Recall that an object will always displace its own volume when placed in a liquid.

Volume of the solid object = Cylinder reading after immersing the object in the water - cylinder reading before immersing the object in the water.

             = 48.1 - 20.4

                  = 27.8 mL

Mass of the solid object = 73.05 g

Density of the object = 73.05/27.8

                                         = 2.63 g/mL

More on density can be found here: brainly.com/question/15164682

#SPJ1

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Answer is: C₃H₃N₃O₃.
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m(CₐHₓNₓ) = 5,214 g.
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n(CO₂) = n(C) =  5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
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n(N) = 0,0607 mol · 2 = 0,121 mol.
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n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.


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C) In[reactant] vs. time

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Taking ln on both sides gives:

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Given that, an experiment to measure the enthalpy change for the reaction of aqueous copper(II) sulfate, CuSO4(aq) and zinc, Zn(s) was carried out in a coffee cup calorimeter; the heat of the reaction in the whole system is calculated to be 2218.34 kJ

Heat of reaction (i.e enthalpy of reaction) is the quantity of heat that is required to be added or removed when a chemical reaction is taken place in order to maintain all of the compounds present at the same temperature.

The formula used to calculate the heat of the reaction can be expressed as follows:

Q = mcΔT

where:

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From the information given:

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ΔT = 91.5° C - 25° C

ΔT = 66.5° C

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\mathbf{= (1 \times \dfrac{50}{1000})\ moles}

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Then;

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By crossing multiplying;

mass of CuSO₄ = number of moles of CuSO₄ ×  molar mass of CuSO₄

mass of CuSO₄ = 0.05 moles  × 159.609 g/moles

mass of CuSO₄ = 7.9805 grams

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Q = mcΔT

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Q = 2218.34 kJ

Therefore, we can conclude that the heat of the reaction is 2218.34 kJ

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