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stich3 [128]
3 years ago
8

What is the volume of a salt crystal measuring 5.44 x 10 ^ - 2 m by 2.5 x 10 ^ - 3 m by 7.9 x10-^ 3 m ?

Chemistry
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

The volume of a salt crystal with the mentioned dimensions is V= 107.44 x 10^-8.

Explanation:

To solve this problem we multiply the length of the sides of the crystal. To calculate this product you have to calculate:

1) (5.44 x 2.5 x 9,9). and

2) (10^-2 x 10^-3 x 10^-3). Here we have the product of powers of equal base so we should sum the exponents.

⇒(10^-2 x 10^-3 x 10^-3) = 10^(-2 + -3 + -3) ⇒ (10^-2 x 10^-3 x 10^-3) = 10^-8.

Finally,

V= (5.44 x 10^-2)x (2.3 x 10^-3) x (7.9 x 10^-3) ⇒ V= 107.44 x 10^-8 m.

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3 years ago
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If a ballon at 3.4 atm at 301k, what is the pressure when cooled to 212k
stepladder [879]

Answer:

2.39 atm

Explanation:

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8 0
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At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

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3 years ago
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A lot of them.

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