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3 years ago

Determining wether the molecule is polar or non polar

Chemistry

1 answer:

Rainbow [258]3 years ago

7 0

Answer:

1) polar

2)polar (not too sure about this one)

3)nonpolar

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In a chemical change

zubka84 [21]

Answer:

the product are different substances frim the starting materails

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3 years ago

Read 2 more answers

If an element A is heavier than element B, then...

Aneli [31]

Answer:

Explanation:

because non metals mostly exist in gaseous and liquid state and since element A is heavier hence its a metal

4 0

3 years ago

What is the mass in grams for a piece of lead that has a volume of 14.8 cubic centimeters? The density of lead is 11.4 g/cc.

Oksi-84 [34.3K]

Answer:

168.72 grams

Explanation:

Density = Mass/Volume

Volume = 14.8

Mass = ?

Density = 11.4

Let x = mass

11.4 = x/14.8

Multiply both sides by 14.8

14.8(11.4 = x/14.8)

x = 168.72

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3 years ago

1 Which scientist first said that "elements are made up of identical atoms

babunello [35]

Answer:

John Dalton

Explanation:

4 0

3 years ago

For the equilibrium

sleet_krkn [62]

Answer:

$\large \boxed{\text{0.091 atm }}$

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

Kc = 8.50 × 10⁻³

n(IBr) = 0.0600 mol

V = 1.0 L

1. Calculate [IBr]

$\text{[IBr]} = \dfrac{\text{0.0600 mol}}{\text{1.0 L}} = \text{0.0600 mol/L}$

2. Set up an ICE table.

$\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} & & \\0 & & 0 & &0.0600 & & \\+x & & +x & &- 2x & & \\x & & x & & 0.0600 - 2x & & \\\end{array}$

3. Calculate [I₂]

$\begin{array}{rcl}K_{\text{c}}&=&\dfrac{\text{[IBr]}^{2}} {\text{[I$_{2}$][Br]$_{2}$}}\\\\8.50 \times 10^{-2}&=&{\dfrac{(0.0600 - 2x)^{2}}{x^{2}}}& &\\\\0.2915x & = &{\dfrac{0.0600 - 2x}{x}}& &\\\\0.2915x & = &0.0600 - 2x\\\\2.2915x & = & 0.0600\\x & = & \textbf{0.026 18 mol/L}\\\end{array}\\$

4. Convert the temperature to kelvins

T = (150 + 273.15) K = 423.15 K

5. Calculate p(I₂)

$\begin{array}{rcl}\\pV & = & nRT\\p & = & cRT\\p & = & \text{0.026 18 mol} \cdot \text{L}^{-1}\times \text{0.082 06 L} \cdot \text{atm} \cdot \text{K}^{-1} \text{mol}^{-1} \times \text{423.15 K}\\& = & \textbf{0.91 atm}\\\end{array}\\\text{The partial pressure of iodine is $\large \boxed{\textbf{0.91 atm}}$}$

6 0

3 years ago