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yawa3891 [41]
3 years ago
5

A piece of gold with a mass of 15.23g and an initial temp of 53 Celsius was dropped into a calorimeter containing 28g of water,

the final temp of the metal and water in the calorimeter was 62 what is the initial temp
Chemistry
1 answer:
Vika [28.1K]3 years ago
6 0

Answer : The initial temperature of water (in ^{0}celcius) = 77.108^{0}C  

Solution : Given,

Mass of gold = 15.23 g

Mass of water = 28 g

Initial temperature of gold = 53^{0}C  

Final temperature of gold = 62^{0}C

Final temperature of water = 62^{0}C

Heat capacity of gold = 12.9J/g^{0}C

Heat capacity of water = 4.18J/g^{0}C

The formula used for calorimetry is,

q = m × c × ΔT

where,

q = heat required

m = mass of an element

c = heat capacity

ΔT = change in temperature

In the calorimetry, the energy as heat lost by the system is equal to the gained by the surroundings.

Now the above formula converted and we get

q_{system}= - q_{surrounding}

m_{system}\times c_{system}\times (T_{final}-T_{initial})_{system}= -[m_{surrounding}\times c_{surrounding}\times (T_{final}-T_{initial})_{surrounding}]

m_{gold}\times c_{gold}\times (T_{final}-T_{initial})_{gold}= -[m_{water}\times c_{water}\times (T_{final}-T_{initial})_{water}]

Now put all the given values in this formula, we get

15.23g\times 12.9J/g^{0}C \times (62^{0}C-53^{0}C )= -[28g\times 4.18J/g^{0}C \times (62^{0}C-T_{\text{ initial of water}})]

By rearranging the terms, we get

T_{\text{ initial water}=77.108^{0}C

Thus the initial temperature of water = 77.108^{0}C  

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If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio
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Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>
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4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o
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<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

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By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = \frac{1}{1}\times 0.0711=0.0711mol of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

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Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g

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\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%

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