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3 years ago

Non polar or polar bonds KRb

1 answer:

8 0

The bond length in a molecule of KBr is measured to be 282.1 pm, compared with a nonpolar covalent radius sum for K and Br of 310,4.

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A frying pan needs a Teflon coating of 1.00 mm in thickness to cover an area of 36.0 square inches. How many ounces of Teflon ar

Otrada [13]

Answer:

m = 0.659 ounce

Explanation:

It is given that,

The thickness of a Teflon coating is, d = 1 mm

Area of the coating, A = 36 inch²

The density of Teflon, d = 0.805 g/mL

We need to find ounces of Teflon are needed.

Firstly, find the volume of the Teflon needed,

1 inch² = 6.4516 cm²

36 inch² = 232.258 cm²

Density,

$\rho=\dfrac{m}{V}$

V is volume of the Teflon needed, V = Ad

So,

$m=\rho V\\\\m=0.805\ g/cm^3\times 232.258\ cm^2 \times 0.1\ cm\\\\m=18.69\ g$

Also, 1 gram = 0.035274 ounce

18.69 gram = 0.659 ounce

So, 0.659 ounces of Teflon are needed.

3 0

3 years ago

What is the molarity of a 0.25-liter solution containing 52 grams of HBr? (1 point)

Pani-rosa [81]

2.6 M hBr

This would be the correct answer.

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3 years ago

What is the change of an atom of one element (atom) to an atom of different element called?

Dahasolnce [82]

Nuclear transmutation is the conversion of one chemical element or isotope into another. In other words, atoms of one element can be changed into atoms of another element by transmutation.

6 0

2 years ago

Read 2 more answers

Anton [14]

Answer:

Weight is the measure of the force of gravity on an object. The mass of an object will never change, but the weight of an item can change based on its location. For example, you may weigh 100 pounds on Earth, but in outer space you would be weightless.

Explanation:

I can't see the options but I hope this helps!

8 0

2 years ago

Read 2 more answers

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago