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3 years ago

Can a liquid boil without the formation of gas bubbles in the liquid? Please explain. G

1 answer:

5 0

I don't think so since boiling is the transition from liquid to gas. The gaseous water needs to rise to the surface and they can't do that without the formation of bubbles.

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Explain how lines latitude and longitude help people find locations on earth.

Bad White [126]

Latitude and longitude are the x and y axis on a grid. When you're trying to find a location on earth, you need to graph it, so you can pinpoint the location. There is no grid if there is no longitude and/or latitude lines. Without one of the lines, it would be very hard to graph, which will be difficult in finding locations on earth. But hey, I'm not too sure. I recommend looking at other people's answers too and searching it up.

8 0

3 years ago

True or False: No matter how plentiful resources are, members of a population will always compete for resources.

Dahasolnce [82]

Answer: true

Explanation:

3 0

4 years ago

Read 2 more answers

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

Why are hydrogen, lithium, and sodium classified as reactive elements?

Sergeu [11.5K]

The alkali metals and hydrogen are reactive because they have only one electron to give in order to complete their valence shell. It is easier to give that one electron so when given the opportunity they will. This means they will react with anything polar or willing to take an electron.

4 0

4 years ago

2 C6H14 + 19 O2 --> 12 CO2 + 14 H2O

Rashid [163]

Answer:

4.06 mol H₂O

Explanation:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

First we convert the given masses of reactants into moles, using their respective molar masses:

250 g O₂ ÷ 32 g/mol = 7.81 mol O₂

50 g C₆H₁₄ ÷ 86 g/mol = 0.58 mol C₆H₁₄

Now we calculate how many O₂ moles would react completely with 0.58 C₆H₁₄ moles, using the stoichiometric coefficients of the reaction:

0.58 mol C₆H₁₄ * $\frac{19molO_2}{2molC_6H_{14}}$ = 5.51 mol O₂

As there are more O₂ moles than required (7.81 vs 5.51), O₂ is the reactant in excess. That means that C₆H₁₄ is the limiting reactant.

Now we can calculate how much water can be formed, using the number of moles of the limiting reactant:

0.58 mol C₆H₁₄ * $\frac{14molH_2O}{2molC_6H_{14}}$ = 4.06 mol H₂O

5 0

3 years ago