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4 years ago

What is the temperature difference if 300 cal are used to warm water

Chemistry

1 answer:

PIT_PIT [208]4 years ago

6 0

Answer:

The temperature difference is $\frac{300}{m} Kkg^{-1}$

Explanation:

The formula that is to used is :

Δ $Q=ms$ Δ $T$

where ,

Δ $Q$ is the heat supplied in calories = 300cal
$m$ is the mass of water taken = m (assumed)
Δ $T$ is the change in temperature
$s$ is the specific heat of water = $1calK^{-1}Kg^{-1}$

ΔT :

$\frac{300}{m*1} \\=\frac{300}{m} Kkg^{-1}$

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4 years ago

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

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Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

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Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

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PIT_PIT [208]

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