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3 years ago

A sample of wood with a mass of 30 g has its

Chemistry

1 answer:

anzhelika [568]3 years ago

5 0

Answer:

0.086J/g°C

Explanation:

The following data were obtained from the question:

Mass (M) = 30g

Initial temperature (T1) = 10°C

Final temperature (T2) = 45°C

Change in temperature (ΔT) = T2 – T1 = 45°C – 10°C = 35°C

Heat (Q) released = 90J

Specific heat capacity (C) of wood =..?

We can obtain the specific heat capacity of the wood as follow:

Q = MCΔT

90 = 30 x C x 35

Divide both side by 30 x 35

C = 90/(30 x 35)

C = 0.086J/g°C.

Therefore, the specific heat capacity of the wood is 0.086J/g°C.

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The standard free energy change for a reaction can be calculated using the equation

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Answer:

a) 1,737 kJ

b) -1,93 kJ

Explanation:

The standard free energy change (ΔG) can be determined using:

ΔG = -nFE

a) For the reaction:

Fumarate²⁻ + CoQH₂ ⇄ succinate²⁻ + CoQ

The transferred electrons are 2, As E=-0,009V:

ΔG = -2×96,5kJ/molV×-0,009V

ΔG = 1,737 kJ

b) For the reaction:

cytc1 (Fe²⁺) + cytc (Fe³⁺) ⇄ cytc1 (Fe³⁺) + cytc (Fe²⁺)

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What kind of reaction is happening when glaze is fired

bija089 [108]

Answer:

Oxidation and Reduction occurs

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3 years ago

Calculate the mass of sodium acetate trihydrate required to make 100. ml of a 0.100 m solution. as part of your preparation for

vova2212 [387]

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3 years ago

During a routine check of the fluoride content of Gotham City\'s water supply, the following results were obtained from replicat

Aleksandr-060686 [28]

Answer:

The mean is $x=0.503\frac{mg}{L}$

The 90% confidence interval is:

$i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}]$

Explanation:

1. First organize the data:

$x_{1}=0.487$

$x_{2}=0.487$

$x_{3}=0.511$

$x_{4}=0.511$

$x_{5}=0.519$

As there are 5 data, the sample size (n) is n=5

2. Calculate the mean x:

The mean is calculated adding up all the data and divide them between the sample size.

$x=\frac{0.511+0.487+0.511+0.487+0.519}{5}$

$x=0.503\frac{mg}{L}$

3. Find 90% confidence interval.

The formula to find the confidence interval is:

$i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})]$ (Eq.1)

where x is the mean, d is the standard deviation and n is the sample size.

And

$1-\alpha=0.90$

$\alpha=0.10$

$\frac{\alpha}{2}=0.05$

$z_{0.05}=1.645$

4. Find the standard deviation

$d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}$

$d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}$

$d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}$

$d=\sqrt{2.24*10^{-4}}$

$d=0.015$

5. Replace values in (Eq.1):

$i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})]$

For the addition:

$i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.514$

For the subtraction:

$i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.492$

The 90% confidence interval is:

$i_{0.90}=[0.492,0.514]$

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3 years ago