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lbvjy [14]
3 years ago
8

How many atoms are in 4 moles of oxygen?

Chemistry
1 answer:
Paha777 [63]3 years ago
5 0

Answer:

There are 6.022 × 10 23 O atoms in a mole of O atoms. There are 6.022 × 10 23 O 2 molecules in a mole of . Since you have 2 oxygen atoms in one molecule, there are 2 × 6.022 × 10 23 O atoms in a mole of . A 'mole' is not short for a 'molecule

Explanation:

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The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

8 0
3 years ago
How many moles of Carbon are in 3.06 g of Carbon
natta225 [31]

Answer:

\boxed {\boxed {\sf 0.255 \ mol \ C }}

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

  • Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Since we are converting 3.06 grams to moles, we multiply by that value.

3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}                      

3.06 * \frac{1 \ mol \ C}{12.011 }    

\frac {3.06}{12.011 } \ mol \ C                                

0.25476646 \ mol \ C

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

  • 0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

0.255 \ mol \ C

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>

3 0
2 years ago
Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation:
tekilochka [14]

Answer:

See explanation

Explanation:

First, let's write the balanced equation again:

2 NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)

Now, we know that the total pressure was 7.76 atm. This total pressure, is the sum of the pressure of water and CO2 like this:

Ptotal = Pwat + PCO2 (1)

This is the dalton's law for partial pressures.

The pressure can be also be relationed with the moles

Ratio of mole = Ratio of pressure

so, taking this in consideration we can say the following:

Pwater/PCO2 = moles water / moles CO2

As the only components exerting pressure are CO2 and Water (Because they are in gas phase), the total pressure can be splitted between the two of them so:

Pwater = Ptotal/2

Pwater = 7.76 / 2 = 3.88 atm

With this pressure, and using the ideal gas equation, we can know the moles of water:

PV = nRT

n = PV/RT     using R = 0.082 L atm / K mol

n = 3.88 * 5 / 0.082 * (160+273)

n = 0.546 moles of water

b) now that we have the moles of water, we can actually know the moles that reacted originally from the sodium carbonate by stechiometry.

2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)    MMCO2 = 84 g/mol

the moles of NaHCO3 initially:

n = 100 / 84

n = 1.19 moles

so, If 1.19 moles of NaHCO3 reacted, and only produces 0.546 moles of water and CO2, then, the remaining moles of NaHCO3 is:

remaining moles = 1.19 - 0.546 = 0.644 moles

therefore the mass remaining:

mCO2 = 0.644 * 84

mCO2 = 54.096 g

c) As it was stated before, only the gaseous components are involved in the pressure, thus, in the kp expression which is:

Kp = Pwater * PCO2

Kp = 3.88 * 3.88

Kp = 15.0544

d) As the total pressure is 7.76 atm and the fact that NaHCO3 is solid, this component is not exerting any pressure in the reaction, as seen in the Kp expression, so it won't matter that if we raise a little the quantity of the reactant, it still has some remaining.

3 0
3 years ago
Hi mate <br><br> Pls help me the 8 one a <br><br> With lots of love: <br> Hareem
Sergeu [11.5K]

Answer:

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3 years ago
Read 2 more answers
what would be the effect on the melting point of a sample if it were not dried completely after filtering the recrystallized sam
jok3333 [9.3K]

The melting point of the sample if it is not dried completely after filtering the recrystallized product will have a broad range and will occur at lower range than the actual value.

What is melting point?

Melting point is the temperature at which the solid form of a given substance changes to the liquid form at atmospheric pressure. It is important because, by using the physical property of a substance the substance can be identified.

The sharp range melting point of the substance indicates the purity of the substance. If the sample is not dried completely after recrystallisation, the melting point will have a broad range.

Therefore, if the sample given is not dried completely, it will be impure and the decreases the melting point of the substance. So the actual melting point of the substance cannot be determined.

To learn more about the melting point click on the given link brainly.com/question/40140

#SPJ4

7 0
1 year ago
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