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OleMash [197]
3 years ago
8

Identify the following elements based on the information provided:

Chemistry
1 answer:
leonid [27]3 years ago
5 0

Answer:

answer

Explanation:

answer

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how many moles of iodine should be added to 750 grams of carbon tetrachloride to prepare a 0.24 m solution
aleksley [76]

Answer:

0.18 mol

Explanation:

Given data

  • Mass of carbon tetrachloride (solvent): 750 g
  • Molality of the solution: 0.24 m
  • Moles of iodine (solute): ?

Step 1: Convert the mass of the solvent to kilograms

We will use the relationship 1 kg = 1,000 g.

750g \times \frac{1kg}{1,000g} =0.750kg

Step 2: Calculate the moles of the solute

The molality is equal to the moles of solute divided by the kilograms of solvent. Then,

m = \frac{moles\ of\ solute }{kilograms\ of\ solvent} \\moles\ of\ solute = m \times kilograms\ of\ solvent = \frac{0.24mol}{kg}  \times 0.750kg = 0.18 mol

6 0
3 years ago
Read 2 more answers
How does the presence of a catalyst affect the enthalpy of a reaction? Group of answer choices 1. It depends on whether you are
Otrada [13]

Answer:

Option 3. The catalyst does not affect the enthalpy change (\Delta H_\text{rxn}) of a reaction.

Explanation:

As its name suggests, the enthalpy change of a reaction (\Delta H_\text{rxn}) is the difference between the enthalpy of the products and the reactants.

On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.

In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change \Delta H_\text{rxn} of the reaction would stay the same.

Refer to an energy profile diagram. Enthalpy change of the reaction \Delta H_\text{rxn} measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.

6 0
3 years ago
ITS DUE TODAY PLSSS HELP John went to the everglades in 1994 why did heh hehehe heheh eheheheheh ehehehe frdee points ezzzzz
sp2606 [1]

Answer:

??? why did you add hehehehe?

Explanation:

6 0
3 years ago
A gas occupies 50.0mL at a standard temp. What volume would it occupy at 335 Celsius with
tino4ka555 [31]

Answer:

V₂ = 111.3 mL

Explanation:

Given data:

Initial volume of gas = 50.0 mL

Initial temperature = standard = 273.15 K

Final volume = ?

Final temperature = 335 °C (335+273.15 = 608.15 K)

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 50.0  mL ×608.15 K / 273.15 k

V₂ = 30407.5 mL.K / 273.15 K

V₂ = 111.3 mL

7 0
3 years ago
Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide. Ca) + H2O --> Ca(OH)2 In a particular
mars1129 [50]

Answer:

Percent yield = 92.5%

Explanation:

The question asks for the percent yield which can be defined as:

\frac{actual yield}{theoretical yield} .100

Where the actual yield is <em>how much product was obtained</em>, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is <em>how much product could be obtained with the given reactants theoretically</em>, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.

1. First lets write the chemical equation reaction correctly and check that it is balanced:

CaO + H₂O → Ca(OH)₂

2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.

For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:

 Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol

 moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles

As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO

So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol

We need to convert this value to grams:

 Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol

Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g

3. Calculate the percent yield:

\frac{actual yield}{theoretical yield} .100

Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %

5 0
3 years ago
Read 2 more answers
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