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3 years ago

If I have 2.5 moles of a gas at a pressure of 7.4 atm and volume of 8.2 liters. What is the temperature

Chemistry

1 answer:

frez [133]3 years ago

4 0

Answer:

295.78K

Explanation:

For this question you can use the ideal gas law, which is

$PV=nRT$

where P is pressure, V is volume, n is moles of substance, R is the ideal gas constant, and T is temperature.

We need to find temperature, so we need to rearrange the equation to solve for T

$T=\frac{PV}{nR}$

We are given n=2.5mol, P=7.4atm, and V=8.2L. We can use the gas constant R=.08206 to find the temperature in kelvin. To do so, plug all of the values into the equation (I'll be excluding units for simplicity)

$T=\frac{(7.4)(8.2)}{(2.5)(.08206)} \\\\ T=295.78$

The temperature is 295.78K. To find it in Celsius, subtract 273.

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Kinetic energy id energy of motion and is equal to 1/2mv^2 where m = mass and v = velocity.

So the answer is a salmon leaping up a waterfall.

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3 years ago

Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)

MrRissso [65]

Answer:

Equilibrium constant expression for $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)$ :

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}$ .

Where

$a_{\mathrm{H_2CO_3}}$ , $a_{\mathrm{H^{+}}}$ , and $a_{\mathrm{CO_3}^{2-}}$ denote the activities of the three species, and
$[\mathrm{H_2CO_3}]$ , $\left[\mathrm{H^{+}}\right]$ , and $\left[\mathrm{CO_3}^{2-}\right]$ denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator. $\rm H_2CO_3\; (aq)$ is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be $\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)$ .

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient " $2$ " on the product side of this reaction. $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ is equivalent to $\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ . The species $\rm H^{+}\, (aq)$ appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

$\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right$ .

That's where the exponent " $2$ " in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}$ .

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous " $(\rm aq)$ " species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}$ .

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3 years ago

Which of the following describes a Brønsted-Lowry base?

Mice21 [21]

__ Brainliest if helped!

Bronsted-Lowry definition of Bases are Proton acceptors,

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3 years ago

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3 years ago

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topjm [15]

Answer:

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3 years ago