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3 years ago

II. Enumerate the following.

Chemistry

1 answer:

Andrej [43]3 years ago

4 0

Answer:

join mjq-yjds-vnp

now go ogle me et

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A sample of octane (c8h18 that has a mass of 0.750 g is burned in a bomb calorimeter. as a result, the temperature of the calori

timurjin [86]

In a closed system, heat should be conserved which means that the heat produced in the calorimeter is equal to the heat released by the combustion reaction. We calculate as follows:

Heat of the combustion reaction = mC(T2-T1)
= 1 (1.50) (41-21)
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3 years ago

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What determines the amount of chemical energy a substance has. 6th grade level please

Galina-37 [17]

<span>The thermodynamic determines the amount of chemical energy a substance has.</span>

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3 years ago

What was the original pressure in a rigid container if the new pressure is 82.5atm and the temperature was raised from 44C to 67

zhenek [66]

Answer: 54 atm

Explanation:

I did 67/82.5 then got 0.8121212121. I them divided 44 by 0.81212121 and got 54.1791044776

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3 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

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2 years ago

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almond37 [142]

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2 years ago