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3 years ago

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You need to measure the length of your driveway. which measuring tool would produce accurate results that you would expect to be

the same as those found by others?
your hand span

a string cut to the length of the tip of your finger to the bottom of your palm

a meter stick with only centimeters

a ruler with millimeters and centimeters

2 answers:

AlekseyPX3 years ago

5 0

Answer:a meter stick with only centimeters

Hope this help:)

11Alexandr11 [23.1K]3 years ago

4 0

I would think a Meter stick since a driveway is big so a string only cut a few inches would not work, mills and centi. are to small so a meter should work best

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Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

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3 years ago

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VladimirAG [237]

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3 years ago

A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure su

Tom [10]

Answer:

$V_2 = 4.87 * 10^3$

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

$PV = nRT$

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

$\frac{PV}{PT} = \frac{nRT}{PT}$

$\frac{V}{T} = \frac{nR}{P}$

Represent $\frac{nR}{P}$ with k

$\frac{V}{T} = k$

$k = \frac{V_1}{T_1} = \frac{V_2}{T_2}$

At this point, we can solve for the required parameter using the following;

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

$\frac{4.5 * 10^3}{298} = \frac{V_2}{323}$

Multiply both sides by 323

$323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323$

$323 * \frac{4.5 * 10^3}{298} = V_2$

$V_2 = 323 * \frac{4.5 * 10^3}{298}$

$V_2 = \frac{323 * 4.5 * 10^3}{298}$

$V_2 = \frac{1453.5 * 10^3}{298}$

$V_2 = 4.87 * 10^3$

Hence, the final volume at 50C is $V_2 = 4.87 * 10^3$

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3 years ago

According to the kinetic molecular theory, when a gas particle hits something what will happen?

Juli2301 [7.4K]

The kinetic model theory states that the particles of gases behave as perfectly elastic spheres, exhibiting elastic collisions, where kinetic energy is conserved, whenever they hit something.

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3 years ago

What is the pH of a 0.1M acid solution? <br><br> A. 1<br> B. 0 <br> C. -1

Degger [83]

Hello!

We have the following data:

[H+] = 0.1 M → $1*10^{-1}$

Formula:

$pH = - log\:[H^+]$

$pH = - log\:[1*10^{-1}]$

$pH = 1 - log\:1$

$pH = 1 - 0$

$\boxed{\boxed{pH = 1}}\end{array}}\qquad\checkmark$

Answer:

A. 1

______________________________

I Hope this helps, greetings ... DexteR!

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3 years ago