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3 years ago

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What are the 4 variables that describe a gas

1 answer:

Phantasy [73]3 years ago

8 0

Pressure, volume, temperature, # moles Pressure, volume and temperature, and moles of gas

Hope that helps!!!!

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N3- ion name pls help!!

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N with a charge of (-3) is nitride

Explanation:

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solubility of the solvent,

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Four ice cubes at exactly 0 ∘c with a total mass of 52.5 g are combined with 160 g of water at 90 ∘c in an insulated container.

horrorfan [7]

Heat gained by ice cubes would be equal to the - heat lost by warm water

The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol

Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J

Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C

In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:

q(ice/water) = - q(warm water)

moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)

50.5 g / 18.0 g/mol = 2.81 mol

2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)

16916 + 211.3T2 = -669.4 T2 + 53555

36639 = 880.7 T2

T2 = 41.6 C

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3 years ago

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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:

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

4 0

3 years ago