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astraxan [27]
3 years ago
12

A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperatu

re must this sample be heated?
Chemistry
1 answer:
Anastaziya [24]3 years ago
7 0

Answer:

The sample will be heated to 808.5 Kelvin

Explanation:

Step 1: Data given

Volume before heating = 2.00L

Temperature before heating = 35.0°C = 308 K

Volume after heating = 5.25 L

Pressure is constant

Step 2: Calculate temperature

V1 / T1 = V2 /T2

⇒ V1 = the initial volume = 2.00 L

⇒ T1 = the initial temperature = 308 K

⇒ V2 = the final volume = 5.25 L

⇒ T2 = The final temperature = TO BE DETERMINED

2.00L / 308.0 = 5.25L / T2

T2 = 5.25/(2.00/308.0)

T2 = 808.5 K

The sample will be heated to 808.5 Kelvin

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Your answer would be 0.024951344877489 but rounding it would be 0.025 moles

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Which of the following is an example of qualitative observation?Select one of the options below as your answer:. A. Joseph is ob
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<h2>Answer : Option C) Joseph is observing the color of the reaction mixture to see whether proteins are present in the given solution.</h2><h3>Explanation :</h3>

An example of qualitative observation is the one where one uses the five senses to identify the changes in the reaction.

Here, when Joseph is studying a reaction mixture he is trying to observe a color change which will confirm that there is proteins present in the reaction mixture or not If there is a color change observed then it will confirm the presence of proteins.

Usually qualitative observations are those which can be easily predicted by using five senses.


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Review this reaction:H2SO4 + NaOH ?.What are the product(s)?Na2SO4H2ONa2SO4 and H2ONaH and SO4OHH3SO5
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3 years ago
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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
How many atoms are found in 3.45g of CO2?
USPshnik [31]

<u>Answer:</u> The number of carbon and oxygen atoms in the given amount of carbon dioxide is 4.72\times 10^{22} and 9.44\times 10^{22} respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of carbon dioxide gas = 3.45 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in above equation, we get:

\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol

1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.

According to mole concept:

1 mole of a compound contains 6.022\time 10^{23} number of molecules

So, 0.0784 moles of carbon dioxide gas will contain 1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22} number of carbon atoms and 2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22} number of oxygen atoms

Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is 4.72\times 10^{22} and 9.44\times 10^{22} respectively

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