All categories

4 years ago

Which of these best explains what happens when ice melts to form liquid water?

1 answer:

6 0

Since the question says “these” I assume it was multiple choice...but anyways, intermolecular forces between some of the water molecules are breaking apart. The H2O is going from a tightly packed solid of a fixed volume to a liquid with molecules moving faster and less bonds, which can take the shape of its container. It requires energy to break a bond, so this is an endothermic process.

You might be interested in

In which of the following sets is the symbol of the element, the number of protons, and the number of electrons given correctly?

Maksim231197 [3]

In, 49 protons, 49 electrons

3 0

4 years ago

Who wrote the first modern chemical textbook?

grin007 [14]

Antoine Lavoisier is the answer

5 0

3 years ago

Read 2 more answers

Help please I put what I thought thank you so much

Alik [6]

Answer:

your answer second once is correct

Explanation:

don't worry! have a good day dear!!..

6 0

3 years ago

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: CO(g) + H2O(g)

taurus [48]

Answer: The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

Explanation:

For the given chemical reaction:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}$ ........(1)

We are given:

$[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M$

$[CO_2]_{eq}=0.40M$

Putting values in above equation, we get:

$K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

$\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M$

When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.

Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial: 0.1 0.1 0.4 0.1

At eqllm: 0.1+x 0.1+x 0.4-x 0.25-x

Putting values in expression 1, we get:

$4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522$

Neglecting the negative value of 'x'

Calculating the concentrations of the species:

Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M

Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

8 0

4 years ago

How many moles are in 2.4g of carbon dioxide (CO2)?

miv72 [106K]

Answer:

$\boxed {\boxed {\sf 0.055 \ mol \ CO_2}}$

Explanation:

To convert form grams to moles, the molar mass must be used. This is the mass (in grams) in 1 mole of a substance.

We can use the values on the Periodic Table. First, find the molar masses of the individual elements: carbon and oxygen.

C: 12.011 g/mol
O: 15.999 g/mol

Check for subscripts. The subscript of 2 after O means there are 2 oxygen atoms, so we have to multiply oxygen's molar mass by 2 before adding.

O₂: 2* (15.999 g/mol)=31.998 g/mol
CO₂: 12.011 g/mol + 31.998 g/mol =40.009 g/mol

Use the molar mass as a ratio.

$\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}$

Multiply by the given number of grams.

$2.4 \ g \ CO_2 *\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}$

Flip the fraction so the grams of carbon dioxide cancel.

$2.4 \ g \ CO_2 *\frac { 1 \ mol \ CO_2}{44.009 \ g\ CO_2}$

$2.4 *\frac { 1 \ mol \ CO_2}{44.009}$

$\frac { 2.4 \ mol \ CO_2}{44.009}= 0.0545342998 \ mol \ CO_2$

The original measurement of grams has 2 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

The ten thousandth place has a 5, so we round the 4 to a 5.

$0.055 \ mol \ CO_2$

2.4 grams of carbon dioxide is about 0.055 moles.

8 0

3 years ago