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3 years ago

How can phisical change what a subtance?

1 answer:

3 0

Physical change can change a substance by affecting the form of a chemical change.
Examples of physical changes:
Something cut (Paper)
Molded (Bread)
Boiled (Water)
Mixed (Berries and strawberries)
HOPE THIS HELPS YOU! ^_^

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At equilibrium at 2500K, [HCl]=0.0625M and [H2]=[Cl2]=0.00450M for the reaction H2+Cl2 ⇌ HCl.

ASHA 777 [7]

Answer:

a. $H_2+Cl_2 \rightleftharpoons 2HCl$

b. K = 192.9

c. Products are favored.

Explanation:

Hello!

a. In this case, according to the unbalanced chemical reaction we need to balance HCl as shown below:

$H_2+Cl_2 \rightleftharpoons 2HCl$

In order to reach 2 hydrogen and chlorine atoms at both sides.

b. Here, given the concentrations at equilibrium and the following equilibrium expression, we have:

$K=\frac{[HCl]^2}{[H_2][Cl_2]}$

Therefore, we plug in the data to obtain:

$K=\frac{(0.0625)^2}{(0.00450)(0.00450)}\\\\K=192.9$

c. Finally, we infer that since K>>1 the forward reaction towards products is favored.

Best regards!

8 0

2 years ago

What elements comprise the largest portion of Earth's atmosphere?

Mrrafil [7]

Answer:

Nitrogen and Oxygen make up 99% of the Earths atmosphere. Then Argon makes up the rest of the atmosphere.

3 0

3 years ago

Bears and salmon have a predator - prey relationship . What would most likely happen to the bear population if the salmon popula

AlexFokin [52]

The population would grow and live easier.

7 0

3 years ago

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Question 1 of 10 What is molarity a measurement of? O A. The volume of liquid containing a dissolved substance B. The grams of a

Nuetrik [128]

Answer:

Explanation:

D. The number of moles of a dissolved substance

7 0

3 years ago

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Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to

vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

$H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As, Molarity = $\frac{\text{moles of solute}}{\text{liters of solution
}}$

1.500 M = $\frac{n}{0.08500 L
}$

n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$ to grams of $NaHCO_3$ . As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

$0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

= 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0

3 years ago